## Category Archives: co-ordinate geometry

### Co-ordinate geometry practice for IITJEE Maths: Ellipses

Problem 1:

Find the locus of the point of intersection of tangents to the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{a^{2}}=1$, which are at right angles.

Solution I:

Any tangent to the ellipse is $y = mx + \sqrt{a^{2}m^{2}+b^{2}}$….call this equation I.

Equation of the tangent perpendicular to this tangent is $y=-\frac{-1}{m}x+\sqrt{\frac{a^{2}}{m^{2}}+b^{2}}$…call this equation II.

The locus of the intersection of tangent lines (i) and (ii) is obtained by eliminating m between these equations. So, squaring and adding them, we get

$(y-mx)^{2}+(my+x)^{2}=a^{2}m^{2}+b^{2}+a^{2}+b^{2}m^{2}$

$\Longrightarrow (1+m^{2})(x^{2}+y^{2})=(1+m^{2})(a^{2}+b^{2})$

$\Longrightarrow x^{2}+y^{2}=a^{2}+b^{2}$

which is a circle with its centre at the centre of the ellipse and radius equal to the length of the line joining the ends of the major and minor axis. This circle is called the director circle of the ellipse.

Problem II:

A tangent to the ellipse $x^{2}+4y^{2}=4$ meets the ellipse $x^{2}+2y^{2}=6$ at P and Q. Prove that the tangents at P and Q of the ellipse $x^{2}+2y^{2}=6$ at right angles.

Solution II:

Let the tangent at $R(2\cos {\theta}, \sin{\theta})$ to the ellipse $x^{2}+4y^{2}=4$ meet the ellipse $x^{2}+2y^{2}=6$ at P and Q.

Let the tangents at P and Q to the second ellipse intersect at the point $S(\alpha,\beta)$. Then, PQ is the chord of contact of the point $S(\alpha,\beta)$ with respect to ellipse two, and so its equation is

$\alpha x + 2\beta y=6$….call this “A”.

PQ is also the tangent at $R(2\cos {\theta}, \sin{\theta})$ to the first ellipse and so the equation can be written as $(2\cos{\theta})x+(4\sin{\theta})y=4$….call this “B”.

Comparing “A” and “B”, we get $\frac{2\cos{\theta}}{\alpha} = \frac{4\sin{\theta}}{2\beta} = \frac{4}{6}$

$\Longrightarrow \cos{\theta}=\frac{\alpha}{3}$ and $\sin{\theta}=\frac{\beta}{3}$

$\Longrightarrow \frac{\alpha^{2}}{9} + \frac{\beta^{2}}{9}=1 \Longrightarrow \alpha^{2}+\beta^{2}=9$

The locus of $S( \alpha, \beta)$ is $x^{2}+y^{2}=9$, or $x^{2}+y^{2}=6+3$, which is the director circle of the second ellipse.

Hence, the tangents at P and Q to the ellipse (ii) are at right angles (by the solution to the previous example).

Problem 3:

Let d be the perpendicular distance from the centre of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}=1$ to the tangent drawn at a point P on the ellipse. If $F_{1}$ and $F_{}{2}$ are the two foci of the ellipse, then show that $(PF_{1}-PF_{2})^{2}=4a^{2}(1-\frac{b^{2}}{d^{2}})$

Solution 3:

Equation of the tangent at the point $P(a\cos {\theta}, b\sin{\theta})$ on the given ellipse is $\frac{x\cos{\theta}}{a} + \frac{y\sin{\theta}}{b}=1$. Thus,

$d= |\frac{-1}{\sqrt{\frac{\cos^{2}{\theta}}{a^{2}}+\frac{\sin^{2}{\theta}}{b^{2}}}}|$

$\Longrightarrow d^{2}=\frac{a^{2}b^{2}}{b^{2}\cos^{2}{\theta}+a^{2}\sin^{2}{\theta}}$

We know $PF_{1}+PF_{2}=2a$

$\Longrightarrow (PF_{1}-PF_{2})^{2}=(PF_{1}+PF_{2})^{2}-4PF_{1}PF_{2}$…call this equation I.

Also, $(PF_{1}.PF_{2})^{2}=[ (a\cos{\theta}-ae)^{2}+(b\sin{\theta})^ {2}].[(a\cos{\theta}+ae)^{2} + (b\sin{\theta})^{2}]$, which in turn equals,

$[a^{2}(\cos{\theta}-e)^{2}+a^{2}(1-e^{2})\sin^{2}{\theta} ]. [a^{2}(\cos{\theta}+e)^{2}+a^{2}(1-e^{2})\sin^{2}{\theta} ]$, that is,

$a^{4}[ (\cos^{2}{\theta}+e^{2}) -2e\cos{\theta}+\sin^{2}{\theta} - e^{2} \sin^{2}{\theta} ]. [ (\cos^{2}{\theta}+e^{2}) + 2e \cos{\theta} + \sin^{2}{\theta} - e^{2}\sin^{2}{\theta}] = a^{4}[ 1-2e\cos{\theta}+e^{2}\cos^{2}{\theta} ] [1+ 2e \cos{\theta} + e^{2}\cos^{2}{\theta} ]$

that is,

$a^{4}[ (1+e^{2}\cos^{2}{\theta})^{2}-4e^{2}\cos^{2}{\theta}] = a^{4}(1-e^{2}\cos^{2}{\theta})^{2}$

$\Longrightarrow PF_{1}. PF_{2}=a^{2}(1-e^{2}\cos^{2}{\theta})$

Now, from I, we get $(PF_{1} - PF_{2})^{2} = 4a^{2}-4a^{2}(1-e^{2}\cos^{2}{\theta}) = 4a^{2}e^{2}\cos^{2}{\theta}$,

also, $1-\frac{b^{2}}{a^{2}} = 1 - \frac{b^{2}\cos^{2}{\theta} + a^{2}\sin^{2}{\theta}}{a^{2}} = \frac{(a^{2}-b^{2})\cos^{2}{\theta}}{a^{2}} = e^{2}\cos^{2}{\theta}$

Hence, $(PF_{1} - PF_{2})^{2} = 4a^{2}(1-\frac{b^{2}}{d^{2}})$

we will continue later,

Cheers,

Nalin Pithwa

### Co-ordinate Geometry problems for IITJEE : equations of median, area of a triangle, and circles

Problem I:

If $A(x_{1}, y_{1})$, $B(x_{1}, y_{1})$ and $C(x_{3}, y_{3})$ are the vertices of a triangle ABC, then prove that the equation of the median through A is given by:

$\left | \begin{array}{ccc} x & y & 1 \\ x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \end{array}\right | + \left | \begin{array}{ccc} x & y & 1 \\ x_{1} & y_{1} & 1 \\ x_{3} & y_{3} & 1 \end{array}\right |=0$

Solution I:

If D is the mid-point of BC, its co-ordinates are $( \frac{x_{2}+x_{3}}{2}, \frac{y_{2}+y_{3}}{2} )$

Therefore, equation of the median AD is $\left | \begin{array}{ccc}x & y & 1 \\ x_{1} & y_{1} & 1\\ \frac{x_{2}+x_{3}}{2} & \frac{y_{2}+y_{3}}{2} & 1 \end{array} \right|=0$, which in turn, implies that,

$\left | \begin{array}{ccc}x & y & 1 \\ x_{1} & y_{1} & 1 \\ x_{2}+x_{3} & y_{2}+y_{3} & 2 \end{array}\right |=0$

Now apply the row transformation $R_{3} \rightarrow 2R_{3}$ to the previous determinant. So, we get

$\left | \begin{array}{ccc}x & y & 1 \\ x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \end{array}\right | + \left | \begin{array}{ccc}x & y & 1 \\ x_{1} & y_{1} & 1 \\ x_{3} & y_{3} & 1 \end{array} \right |=0$, using the sum property of determinants.

Hence, the proof.

Problem 2:

If $\triangle_{1}$ is the area of the triangle with vertices $(0,0), (a\tan {\alpha},b\cot{\alpha}), (a\sin{\alpha}, b\cos {\alpha})$, and $\triangle_{2}$ is the area of the triangle with vertices $(a,b)$, $(a\sec^{2}{\alpha}, b\csc^{2}{\alpha})$, and $(a+a\sin^{2}{\alpha}, b+b\cos^{2}{\alpha})$, and $\triangle_{3}$ is the area of the triangle with vertices $( 0, 0)$, $( a\tan{\alpha}, -b\cos{\alpha})$, $(a\sin{\alpha},b\cos{\alpha})$. Then, prove that there is no value of $\alpha$ for which the areas of triangles, $\triangle_{1}$, $\triangle_{2}$ and $\triangle_{3}$ are in GP.

Solution 2:

We have $\triangle_{1}=\frac{1}{2}|\left | \begin{array}{ccc}0 & 0 & 1 \\ a\tan{\alpha} & b\cot {\alpha} & 1 \\ a \sin{\alpha} & b\cos{\alpha} & 1 \end{array}\right ||=\frac{1}{2}ab|\sin{\alpha}-\cos{\alpha}|$, and

$\triangle_{2}=\frac{1}{2}|\left | \begin{array}{ccc}a & b & 1 \\ a\sec^{2}{\alpha} & b\csc^{2}{\alpha} & 1 \\ a + a\sin^{2}{\alpha} & b + b\cos^{2}{\alpha} & 1 \end{array} \right | |$.

Applying the following column transformations to the above determinant, $C_{1} \rightarrow -aC_{3}$ and $C_{2}-bC_{3}$, we get

$\triangle_{2}=\frac{1}{2}ab\left | \begin{array}{ccc}0 & 0 & 1 \\ \tan^{2}{\alpha} & \cot^{2}{\alpha} & 1 \\ \sin^{2}{\alpha} & \cos^{2}{\alpha} & 1 \end{array}\right | = \frac{1}{2}ab(\sin^{2}{\alpha}-\cos^{2}{\alpha})$ and $\triangle_{3}=\frac{1}{2}|\left | \begin{array}{ccc} 0 & 0 & 1 \\ a\tan{\alpha} & -b\cot{\alpha} & 1 \\ a\sin{\alpha} & b\cos{\alpha} & 1 \end{array} \right | |=\frac{1}{2}ab |\sin {\alpha}+\cos{\alpha}|$

so that $\triangle_{1}\triangle_{3}=\frac{1}{2}ab\triangle_{2}$.

Now, $\triangle_{1}$, $\triangle_{2}$ and $\triangle_{3}$ are in GP, if $\triangle_{1}\triangle_{3}=\triangle_{2}^{2} \Longrightarrow \frac{1}{2}ab\triangle_{2}=\triangle_{2}^{2} \Longrightarrow \triangle_{2}=\frac{1}{2}ab$

$\Longrightarrow \triangle_{2}=\frac{1}{2}ab(\sin^{2}{\alpha}-\cos^{2}{\alpha})=\frac{1}{2}ab \Longrightarrow (\sin^{2}{\alpha}-\cos^{2}{alpha})=1$, that is,

$\alpha = (2m+1)\pi/2$, where $m \in Z$. But, for this value of $\alpha$, the vertices of the given triangles are not defined. Hence, $\triangle_{1}$, and $\triangle_{2}$ and $\triangle_{3}$ cannot be in GP for any value of $\alpha$.

Problem 3:

Two points P and Q are taken on the line joining the points $A(0,0)$ and $B(3a,0)$ such that $AP=PQ=QB$. Circles are drawn on AP, PQ, and QB as diameters. The locus of the point S, the sum of the squares of the length of the tangents from which to the three circles is equal to $b^{2}$, is

(a) $x^{2}+y^{2}-3ax+2a^{2}-b^{2}=0$

(b) $3(x^{2}+y^{2})-9ax+8a^{2}-b^{2}=0$

(c) $x^{2}+y^{2}-5ax+6a^{2}-b^{2}=0$

(d) $x^{2}+y^{2}-ax-b^{2}=0$.

Ans. b.

Solution 3:

Since $AP=PQ=QB$, the co-ordinates of P are $(a,0)$ and of Q are $(2a,0)$, equations of the circles on AP, PQ, and QB as diameters are respectively.

So, we get

$(x-0)(x-a)+y^{2}=0$

$(x-a)(x-2a)+y^{2}=0$

$(x-2a)(x-3a)+y^{2}=0$

So, if $(h,k)$ be any point of the locus, then $3(h^{2}+k^{2})-9ah+8a^{2}=b^{2}$.

So, the required locus of $(h,k)$ is $3(x^{2}+y^{2})-9ax+8a^{2}-b^{2}=0$.

More later,

Nalin Pithwa.

### Co-ordinate Geometry : IITJEE Mains practice: some random problems again

Problem 1:

The line $Ax+By+C=0$ cuts the circle $x^{2}+y^{2}+ax+by+c=0$ at P and Q. The line $A^{'}x+B^{'}y+C^{'}=0$ cuts the circle $x^{2}+y^{2}+a^{'}x+b^{'}y+c^{'}=0$ at R and S. If P, Q, R and S are concyclic, prove that

$\left | \begin{array}{ccc} a-a^{'} & b-b^{'} & c-c^{'} \\ A & B & C \\ A^{'} & B^{'} & C^{'} \end{array}\right |=0$.

Solution I;

An equation of a circle through P and Q is $x^{2}+y^{2}+ax+by+c +\lambda (Ax+By+C)=0$…call this equation I.

And, an equation of a circle through R and S is $x^{2}+y^{2}+a^{'}x+b^{'}y + c^{'}+\mu (A^{'}x+B^{'}y+C^{'})=0$…call this equation II.

If P, Q, R and S are concyclic, then I and II represent the same circle for same values of $\lambda$ and $\mu$.

$\Longrightarrow a+ \lambda A=a^{'}+\mu A^{'}$ or $a-a^{'} + \lambda A - \mu A^{'}=0$

so also,

$b + \lambda B = b^{'} + \mu B^{'}$ or $b-b^{'}+\lambda B - \mu B^{'}=0$

$c + \lambda C = c^{'} + \mu C^{'}$ or $c-c^{'} + \lambda C - \mu C^{'}=0$.

Eliminating $\lambda$ and $\mu$, we get the following:

$\left | \begin{array}{ccc} a-a^{'} & A & -A^{'}\\ b-b^{'} & B & -B^{'}\\ c-c^{'} & C & -C^{'} \end{array} \right |=0$, that is,

$\left | \begin{array}{ccc} a-a^{'} & b-b^{'} & c-c^{'} \\ A & B & C \\ A^{'} & B^{'} & C^{'} \end{array}\right |=0$

Problem II:

A straight line is such that the algebraic sum of the perpendiculars falling upon it from any number of fixed points is zero. Show that it always passes through a fixed point.

Solution II:

Let $(x_{i},y_{i})$ where $i=1,2,3,\ldots$ be n fixed points. Let $ax+by+c=0$ be the given line. Thus, as per given hypthesis, we have

$\sum_{i=1}^{n}\frac{ax{i}+by_{i}+c}{\sqrt{(a^{2}+b^{2})}}=0 \Longrightarrow a\sum_{i=1}^{n}x_{i}+b\sum_{i=1}^{n}y_{i}+nc=0 \Longrightarrow a\overline{x}+b\overline{y}+c=0$ where $\overline{x}=\frac{1}{n}\sum_{i=1}^{n}x_{i}$ and $\overline{y}=\frac{}{}\sum_{i=1}^{n}y_{i}$

which shows that the given line passes through the fixed point $(\frac{1}{n}\sum_{i=1}^{n}x_{i}, \frac{1}{n}\sum_{i=1}^{n}y_{i})$.

Problem III:

The straight lines $L \equiv ax+by+c=0$ and $L_{1} \equiv a_{1}x+b_{1}y+c_{1}=0$ are intersecting. Find the straight line $L_{2}$ such that L is the bisector of the angle between $L_{1}$ and $L_{2}$.

Solution III:

Let the equation of the line $L_{2}$ be $L_{1}+ \lambda L=0 \Longrightarrow (a_{1}+\lambda a)x+(b_{1}+\lambda b)y+\lambda c=0$ where the slopes of $L_{2}, L, L_{1}$ are respectively

$-\frac{a_{1}+\lambda a}{b_{1}+\lambda b}, \frac{-a}{b}, -\frac{a_{1}}{b_{1}}$.

Since L is the bisector of the angle between $L_{2}$ and $L_{1}$ we have

$\frac{-(\frac{a_{1}+\lambda a}{b_{1}+\lambda b})+\frac{a}{b}}{1+\frac{a(a_{1}+\lambda a)}{b(b_{1}+\lambda b)}}=\frac{-\frac{a}{b}+\frac{a_{1}}{b_{1}}}{1+\frac{aa_{1}}{bb_{1}}}$

$\Longrightarrow \frac{-b(a_{1}+\lambda a)+a(a_{1}+\lambda b)}{b(b_{1}+\lambda b)+a(a_{1}+\lambda a)}=-\frac{ab_{1}-a_{1}b}{aa_{1}+bb_{1}}$

$\Longrightarrow \frac{ab_{1}-a_{1}b}{\lambda (a^{2}+b^{2})+aa_{1}+bb_{1}} = -\frac{ab_{1}-a_{1}b}{aa_{1}+bb_{1}}$

$\Longrightarrow \lambda = - \frac{2(aa_{1}+bb_{1})}{a^{2}+b^{2}}$

Hence, the equation of the required line $L_{1}$ is $(a^{2}+b^{2})(a_{1}x+b_{1}y+c_{1})=2(aa_{1}+bb_{1})(ax+by+c)$.

Problem IV:

If a, b are real numbers and $c>0$, find the locus represented by $|ay-bx|=c\sqrt{(x-a)^{2}+(y-b)^{2}}$.

PS: Please draw a right angled triangle PMA, with right angle at M, and P being $(x,y)$ and A being $(a,b)$.

Solution IV:

Let $x=a+r\cos {\theta}$ and $y=b+r\sin {\theta}$, then the given equation becomes $a\sin {\theta}-b\cos {\theta}=c$.

$\Longrightarrow r\sin{(\theta-\alpha)}=c$ where $r=\sqrt{a^{2}+b^{2}}$ and $\tan {(\alpha)}=\frac{b}{a}$ which is the slope of $ay-bx$, which in turn implies $\frac{c}{r}=\sin (\theta -\alpha) \leq 1$

$\Longrightarrow c \leq r$, or $c \leq \sqrt{a^{2}+b^{2}}$. The given equation now becomes

$\frac{|ay-bx|}{\sqrt{a^{2}+b^{2}}}=\frac{c}{\sqrt{a^{2}+b^{2}}}\sqrt{(x-a)^{2}+(y-b)^{2}}$….call this as relation I.

If M is the foot of the perpendicular from a point P(x,y) on the line $ay-bx=0$ and A is the point $(a,b)$ which clearly lies on this line, then from relation I, we have

$\frac{PM}{PA}=\frac{c}{\sqrt{a^{2}+b^{2}}}=\sin {(\theta - \alpha)}$. Hence, the locus of P is a straight line through the point $(a,b)$ inclined at an angle $\arcsin {\frac{c}{\sqrt{a^{2}+b^{2}}}}$ with the line $ay-bx=0$.

Problem V:

Find the co-ordinates of the orthocentre of the triangle formed by the lines $y=0$ and $(1+t)x-ty+t(1+t)=0$ and $(1+u)x-uy+u(1+u)=0$, where $t \neq u$, and show that for all values of t and u, the orthocentre lies on the line $x+y=0$.

Solution V:

Let the equation of the side BC be $y=0$. Then, the coordinates of B and C are $(-t,0)$ and $(-u,0)$, respectively, where $(1+t)x-ty+t(1+t)=0$ and $(1+u)x-uy+u(1+u)=0$ are equations of AB and AC, respectively.

PS: Please draw the diagram on your own for a better understanding of the solution presented.

Now, equation of BE is $y={\frac{-u}{1+u}}(x+t)$…let us call this equaiton I.

And, equation of CF is $y=\frac{-t}{1+t}(x+u)$…let us call this equation II.

Solving I and II, we get the following:

$x(\frac{u}{1+u}-\frac{t}{1+t})=\frac{tu}{1+t}-\frac{tu}{1+u}$, which in turn implies that

$x=tu$ and $y=-tu$, so that the orthocentre is the point $(tu,-tu)$ which lies on the line $x+y=0$.

Cheers,

Nalin Pithwa

### Some random problems/solutions in Coordinate Geometry II: IITJEE mathematics training

Question I:

Find the equation of the tangent to the circle $x^{2}-y^{2}-4x-8y+16=0$ at the point $(2+\sqrt{3},3)$. If the circle rolls up along this tangent by 2 units, find its equation in the new position.

Solution I:

The centre $C_{1}$ of the given circle is $(2,4)$ and its radius is 2. Equation of the tangent at $A(2+\sqrt{3},3)$ to the circle is

$x(2+\sqrt{3})+3y-2(x+2+\sqrt{3})-4(y+3)+16=0$

or $\sqrt{3}x-y-2\sqrt{3}=0$.

The slope of this line is $\sqrt{3}$ showing that it makes an angle of 60 degrees with the x-axis. After the circle rolls up along the tangent at A through a distance 2 units, its centre moves from $C_{1}$ to $C_{2}$. We now find the co-ordinates of $C_{2}$. Since $C_{1}C_{2}$ is parallel to the tangent at A and it passes through $C_{1}$$(2,4)$, its equation is $\frac{x-2}{\cos {\theta}} = \frac{y-4}{\sin {\theta}}$, where $\theta=60 \deg$; $C_{2}$ being at a distance 2 units on this line from $C_{1}$; its co-ordinates are

$(2\cos {\theta}+2, 2\sin{\theta}+4)$, that is, $(3, 4+\sqrt{3})$.

Hence, the equation of the circle in the new position is

$(x-3)^{2}+(y-(4+\sqrt{3})^{2})=2^{2}$, which in turn implies that

$x^{2}+y^{2}-6x-2(4+\sqrt{3})y+8(3+\sqrt{3})=0$.

Question 2:

A triangle has two of its sides along the axes, its third side touches the circle $x^{2}+y^{2}-2ax-2ay+a^{2}=0$. Prove that the locus of the circumcentre of the triangle is

$a^{2}-2a(x+y)+2xy=0$.

Solution 2:

The given circle has its centre at $C(a,a)$ and its radius is a so that it touches both the axes along which lie the two sides of the triangle. Let the third side be $\frac{x}{p} + \frac{y}{p}=1$.

So that A is $(p,a)$ and B is $(a,q)$ and the line AB touches the given circle. Since $\angle AOB$ is a right angle, AB is diameter of the circumcentre of the triangle AOB. So, the circumcentre $P(h,k)$ of the triangle AOB is the mid-point of AB,

that is, $2h=p$, $2k=q$.

Now, the equation of AB is $\frac{x}{p} + \frac{y}{q}=1$, which touches the given circle,

$\frac{a(p+q)-pq}{\sqrt{p^{2}+q^{2}}}=a$

$\Longrightarrow a^{2}(p+q)^{2}+p^{2}-2apq(p+q)=a^{2}(p^{2}+q^{2})$

$\Longrightarrow 2a^{2}-2a(p+q)+pq=0$

$2a^{2}-2a(2h+2k)+2h-2k=0$.

Hence, the locus of $P(h,k)$ is $a^{2}-2a(x+y)+2xy=0$.

Question 3:

A circle of radius 2 units rolls on the outerside of the circle $x^{2}+y^{2}+4x=0$, touching it externally. Find the locus of the centre of this outside circle. Also, find the equations of the common tangents of these two circles when the line joining the centres of the two circles make an angle of 60 degrees with x-axis.

Solution 3:

The centre C of the given circle is $(-2,0)$ and its radius is 2. Let $P(h,k)$ be the centre of the outer circle touching the given circle externally then $CP=2+2=4$, which in turn implies,

$(h+2)^{2}+k^{2}=4^{2}$

So, the locus of P is $(x+2)^{2}+y^{2}=16$, or $x^{2}+y^{2}+4x-12=0$.

Since the two circles touch each other externally,, there are 3 common tangents to these circles.

One will be perpendicular to the line joining the centres and the other two will be parallel to the line joining the centres as the radii of the two circles are equal, co-ordinates of P are given by

$\frac{h+2}{\cos{60 \deg}} = \frac{k-0}{\sin{60 \deg}}=4 \Longrightarrow h=0, k=2\sqrt{3}$,

co-ordinates of M, the mid-point of CP is $(-1,\sqrt{3})$.

Hence, the equation of the common tangent perpendicular to CP is

$y-\sqrt{3}=-\frac{1}{\sqrt{3}}(x+1) \Longrightarrow x+\sqrt{3}y-2=0$.

Let the equation of the common tangent parallel to CP be $\sqrt{3}x-y+\lambda=0$.

Since it touches the given circle $\frac{}{}= \pm 2 \Longrightarrow \lambda = 2\sqrt{3} \pm 4$.

Hence, the other common tangents are $\sqrt{3}x -y \pm 4 + 2\sqrt{3}=0$.

Question 4:

If $S=0$ and $S^{'}=0$ are the equations of two circles with radii r and $r^{'}$ respectively, then show that the circles $\frac{S}{r} \pm \frac{S^{'}}{r}=0$ cut orthogonally.

Solution 4:

Let the line of centres of the given circle be taken as the x-axis and its mid-point as the origin…Note this is the key simplifying assumption.

If the distance between the centres is 2a, the co-ordinates of the centre are $(a,0)$ and $(-a,0)$. Hence, we get the following:

$S \equiv (x-a)^{2}+y^{2}-r^{2}=0$, that is,

$S^{'} \equiv x^{2}+y^{2}-2ax +a^{2}-r^{2}=0$

and $S^{'} \equiv x^{2}+y^{2}+2ax+a^{2}-(r^{'})^{2}=0$ so that $\frac{}{} + \frac{}{}=0 \Longrightarrow Sr^{'}+S^{'}r^{'}=0$, that is,

$\Longrightarrow (r+r^{'})(x^{2}+y^{2}+a^{2})-2ax(r^{'}-r)-rr^{'}(r+r^{'})=0$

$\Longrightarrow x^{2}+y^{2}-2a\frac{r^{'}-r}{r^{'}+r}x + a^{2}-rr^{'}=0$…call this I.

and $\frac{S}{r} - \frac{S^{'}}{r^{'}}=0$ and in turn $\Longrightarrow x^{2}+y^{2}-2a\frac{r^{'}+r}{r^{'}-r}+a^{2}+rr^{'}=0$…call this II.

Now, since $2(\{ -a\frac{r^{'}-r}{r^{'}+r}\})(\{-a\frac{r^{'}+r}{r^{'}-r} \})+ 2\times 0 \times 0 =2a^{2}=(a^{2}-rr^{'}) +(a^{2}+rr^{'})$.

The circles I and II intersect orthogonally.

Question 5:

Let P, Q, R, S be the centres of the four circles each of which is cut by a fixed circle orthogonally. If $t_{1}^{2}$, $t_{2}^{2}$, $t_{3}^{2}$, $t_{4}^{2}$ be the squares of the lengths of the tangents to the four circles from a point in their plane, then prove that

$t_{1}^{2}\Delta QRS +t_{2}^{2}\Delta RSP + t_{3}^{2}\Delta SPQ + t_{4}^{2}\Delta PQR=0$

Solution 5:

Let the equations of the four circles be

$x^{2}+y^{2}+2g_{i}x+2f_{i}y+c_{i}=0$, $i=1,2,3,4$, then centres of these circles are as follows:

$P(-g_{1},-f_{1})$, $Q(-g_{2},-f_{2})$, $R(-g_{3},-f_{3})$, and $S(-g_{4},-f_{4})$

Let the fixed point in the plane be taken as the origin, then $t_{1}^{2}=c_{1}$, $t_{2}^{2}=c_{2}$, $t_{3}^{2}=c_{3}$ and $t_{4}^{2}=c_{4}$. Let the equation of the fixed circle cutting the four circles orthogonally be

$x^{2}+y^{2}+2gx + 2fy +c=0$, then $2gg_{1}+2ff_{1}=c+c_{1}=c+t_{1}^{2}$, or

we get the following:

$2gg_{i}+2ff_{i}-c-t_{i}^{2}=0$, for $i=1,2,3,4$.

Eliminating the unknowns g, f, c we get

$\left| \begin{array}{cccc} 2g_{1} & 2f_{1} & -1 & -t_{1}^{2} \\ 2g_{2} & 2f_{2} & -1 & -t_{2}^{2}\\ 2g_{3} & 2f_{3} & -1 & -t_{3}^{2}\\ 2g_{4} & 2f_{4} & -1 & -t_{4}^{2} \end{array} \right|$

or, $t_{1}^{2}|D_{1}|-t_{2}^{2}|D_{2}|+t_{3}^{2}|D_{3}|-t_{4}^{2}|D_{4}|=0$

where $|D_{1}|= \left| \begin{array}{ccc} g_{2} & f_{2} & 1\\ g_{1} & f_{1} & 1 \\ g_{4} & f_{4} & 1 \end{array} \right |=2\Delta QRS$,

and $|D_{2}|=2\Delta PRS$, $|D_{3}|=2\Delta PQS$ and $|D_{4}|=2\Delta PQR$

Hence, we get the following:

$t_{1}^{2}\Delta QRS + t_{2}^{2}\Delta RSP + t_{3}^{2}SPQ + t_{4}^{2}PQR=0$.

Homework Quiz Coordinate Geometry:

1. OAB is any chord of a circle which passes through O, a point in the plane of the circle and meets it in points A and B. A point P is taken on this chord such that OP is (i) arithmetic mean (ii) geometric mean of OA and OB. Prove that the locus of P in either case is a circle. Determine the circle.
2. Let $2x^{2}+y^{2}-3xy=0$ be the equation of a pair of tangents drawn from the origin O to a circle of radius 3 with centre in the first quadrant. If A is one of the points of contact, find the length OA.
3. Let P, Q and R be the centres and $r_{1}, r_{2}, r_{3}$ are the radii respectively of three coaxial circles. Show that $r_{1}^{2}QR + r_{2}^{2}RP + r_{3}^{2}PQ=-PQ. QR. RP$
4. If ABC be any triangle and $A^{'}B^{'}C^{'}$ be the triangle formed by the polars of the points A, B, C with respect to a circle, so that $B^{'}C^{'}$ is the polar of A; $C^{'}A^{'}$ is the polar of B and $A^{'}B^{'}$ is the polar of C. Prove that the lines $AA^{'}$, $BB^{'}$ and $CC^{'}$ meet in a point.

That’s all, folks !

Nalin Pithwa.

### Some random sample problems-solutions in Coordinate Geometry: I: IITJEE Mains Maths tutorials

Question I:

The point $(4,1)$ undergoes the following transformations, successively:

a) reflection about the line $y=x$.

b) translation through a distance 2 units along the positive directions of the x-axis.

c) rotation through an angle of $\pi/4$ about the origin in the anticlockwise direction.

d) reflection about $x=0$.

Hint: draw the diagrams at very step!

Ans: $(1/\sqrt{2}, 7/\sqrt{2})$

Question 2:

$A_{1}, A_{2}, A_{3}, \ldots, A_{n}$ are n points in a plane whose co-ordinates are $(x_{1}, y_{1})$, $(x_{2},y_{2})$, $\ldots$, $(x_{n},y_{n})$ respectively. $A_{1}$, $A_{2}$ is bisected at the point $G_{1}$, $G_{1}A_{3}$ is divided in the ratio 1:2 at $G_{2}$, $G_{2}A_{4}$ is divided in the ratio $1:3$ at $G_{3}$, $G_{3}A_{3}$ is divided in the ratio $1:4$ at $G_{4}$ and so on until all n points are exhausted. Show that the co-ordinates of the final point so obtained are

$(\frac{1}{n}(x_{1}+x_{2}+ \ldots + x_{n}) , \frac{1}{n}(y_{1}+y_{2}+ \ldots + y_{n}) )$.

Solution 2:

The co-ordinates of $G_{1}$ are $(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})$.

Now, $G_{2}$ divides $G_{1}A_{3}$ in the ratio $1:2$. Hence, the co-ordinates of $G_{2}$ are

$( \frac{1}{3}(\frac{2(x_{1}+x_{2})}{2}+x_{3}), \frac{1}{3}(\frac{3(y_{1}+y_{2})}{2}+y_{3}))$, or $(\frac{x_{1}+x_{2}+x_{3}}{3},\frac{y_{1}+y_{2}+y_{3}}{3})$.

Again, $G_{3}$ divides $G_{2}A_{4}$ in the ratio $1:4$. Therefore, the co-ordinates of $G_{3}$ are $(\frac{1}{4}(\frac{3(x_{1}+x_{2}+x_{3})}{3}+x_{4}) ,\frac{1}{4}(\frac{3(y_{1}+y_{2}+y_{3})}{3}+y_{4}) )$, or

$( \frac{x_{1}+x_{2}+x_{3}+x_{4}}{4},\frac{y_{1}+y_{2}+y_{3}+y_{4}}{4} )$.

Proceeding in this manner,we can show that the coordinates of the final point obtained will be

$(\frac{1}{n}(x_{1}+x_{2}+x_{3}+\ldots + x_{n}),\frac{1}{n}(y_{1}+y_{2}+y_{3}+\ldots + y_{n}))$.

Remark: For a rigorous proof, prove the above by mathematical induction.

Question 3:

A line L intersects the three sides BC, CA, and AB of a triangle ABC at P, Q and R, respectively. Show that $\frac{BP}{PC}.\frac{CQ}{QA}.\frac{AR}{RB}=-1$

Solution 3:

Let $A(x_{1},y_{1})$, $B(x_{2},y_{2})$, and $C(x_{3},y_{3})$ be the vertices of $\triangle ABC$, and let $lx+my+n=0$ be equation of the line L. If P divides BC in the ratio $\lambda:1$, then the coordinates of P are $(\frac{\lambda x_{3}+x_{2}}{\lambda + 1} ,\frac{\lambda y_{3}+y_{2}}{\lambda + 1})$.

Also, as P lies on L, we have $l(\frac{\lambda x_{3}+x_{2}}{\lambda + 1})+m(\frac{\lambda y_{3}+y_{2}}{\lambda + 1})+n=0$

$\Longrightarrow \frac{lx_{2}+my_{2}+n}{lx_{3}+my_{3}+n}=\lambda=\frac{BP}{PC}$…..call this relation I.

Similarly, we can obtain $\frac{CQ}{QA}=-\frac{lx_{3}+my_{3}+n}{lx_{1}+my_{1}+n}$….call this relation II.

and so, also, we can prove that $\frac{AR}{RB}=-\frac{lx_{1}+my_{1}+n}{lx_{2}+my_{2}+n}$…call this III.

Multiplying, I, II and III, we get the desired result.

The above is the famous Menelaus’s theorem of plane geometry proved with elementary tools of co-ordinate geometry. As a homework quiz, try proving the equally famous Ceva’s theorem of plane geometry with elementary tools of co-ordinate geometry.

Question 4:

A triangle has the lines $y=m_{1}x$ and $y=m_{2}x$ as two of its sides, with $m_{1}$ and $m_{2}$ being roots of the equation $bx^{2}+2hx+a=0$. If $H(a,b)$ is the orthocentre of the triangle, show that the equation of the third side is $(a+b)(ax+by)=ab(a+b-2h)$.

Solution 4:

Since the given lines intersect at the origin, one of the triangle lies at the origin O(0,0). Let OA and OB be the given lines $y=m_{1}x$ and $y=m_{2}x$, respectively. Let the equation of AB be $lx+my=1$. Now, as OH is perpendicular to AB, we have

$\frac{b}{a}=\frac{m}{l}$, $\Longrightarrow \frac{l}{a}=\frac{m}{b}=k$, say…call this equation I

Also, the coordinates of A and B are respectively,

$(\frac{1}{l+mm_{1}}, \frac{m_{1}}{l+mm_{1}})$ and $(\frac{1}{l+mm_{2}} , \frac{m_{2}}{l+mm_{2}})$

Therefore, the equation of AB is

$(y-\frac{m_{1}}{l+mm_{1}})=-\frac{1}{m_{2}}(x-\frac{1}{l+mm_{1}})$

or $x+m_{2}y=\frac{1+m_{1}m_{2}}{1+mm_{1}}$…call this II.

Similarly, the equation of BH is $x+m_{1}y=\frac{1+m_{1}m_{2}}{1+mm_{2}}$….call this III.

Solving II and III, we get the coordinates of H. Subtracting III from II, we get

$y=\frac{(1+m_{1}m_{2})m}{l^{2}+lm(m_{1}+m_{2})+m^{2}m_{1}m_{2}}$

Since $m_{1}$ and $m_{2}$ are the roots of the equation $bx^{2}+2hx+a=0$, we have $m_{1}+m_{2}=-\frac{2h}{b}$ and $m_{1}m_{2}=a/b$.

$\Longrightarrow y=\frac{(a+b)m}{bl^{2}-2hlm+am^{2}} \Longrightarrow \frac{m}{b}=\frac{bl^{2}-2hlm+am^{2}}{a+b}$ because y=b for H.

$\Longrightarrow k=\frac{k^{2}(ba^{2}-2hab+ab^{2})}{a+b} \Longrightarrow k=\frac{a+b}{ab(a-2h+b)}$.

Hence, the equation of AB is

$ax+by=\frac{1}{k}=\frac{ab(a+b-2h)}{a+b}$

$\Longrightarrow (a+b)(ax+by)=ab(a+b-2h)$

More later,

Nalin Pithwa.

### Circles and System of Circles: IITJEE Mains: some solved problems I

Part I: Multiple Choice Questions:

Example 1:

Locus of the mid-points of the chords of the circle $x^{2}+y^{2}=4$ which subtend a right angle at the centre is (a) $x+y=2$ (b) $x^{2}+y^{2}=1$ (c) $x^{2}+y^{2}=2$ (d) $x-y=0$

Solution 1:

Let O be the centre of the circle $x^{2}+y^{2}=4$, and let AB be any chord of this circle, so that $\angle AOB=\pi /2$. Let $M(h,x)$ be the mid-point of AB. Then, OM is perpendicular to AB. Hence, $(AB)^{2}=(OA)^{2}+(AM)^{2}=4-2=2 \Longrightarrow h^{2}+k^{2}=2$. Therefore, the locus of $(h,k)$ is $x^{2}+y^{2}=2$.

Example 2:

If the equation of one tangent to the circle with centre at $(2,-1)$ from the origin is $3x+y=0$, then the equation of the other tangent through the origin is (a) $3x-y=0$ (b) $x+3y=0$ (c) $x-3y=0$ (d) $x+2y=0$.

Solution 2:

Since $3x+y=0$ touches the given circle, its radius equals the length of the perpendicular from the centre $(2,-1)$ to the line $3x+y=0$. That is,

$r= |\frac{6-1}{\sqrt{9+1}}|=\frac{5}{\sqrt{10}}$.

Let $y=mx$ be the equation of the other tangent to the circle from the origin. Then,

$|\frac{2m+1}{\sqrt{1+m^{2}}}|=\frac{5}{\sqrt{10}}=25(1+m^{2})=10(2m+1)^{2} \Longrightarrow 3m^{2}+8m-3=0$, which gives two values of m and hence, the slopes of two tangents from the origin, with the product of the slopes being -1. Since the slope of the given tangent is -3, that of the required tangent is 1/3, and hence, its equation is $x-3y=0$.

Example 3.

A variable chord is drawn through the origin to the circle $x^{2}+y^{2}-2ax=0$. The locus of the centre of the circle drawn on this chord as diameter is (a) $x^{2}+y^{2}+ax=0$ (b) $x^{2}+y^{2}+ay=0$ (c) $x^{2}+y^{2}-ax=0$ (d) $x^{2}+ y^{2}-ay=0$.

Solution 3:

Let $(h,k)$ be the centre of the required circle. Then, $(h,k)$ being the mid-point of the chord of the given circle, its equation is $hx+ky-a(x+h)=h^{2}+k^{2}-2ah$.

Since it passes through the origin, we have $-ah=h^{2}+k^{2}-2ah \Longrightarrow h^{2}+k^{2}-ah=0$.

Hence, locus of $(h,k)$ is $x^{2}+y^{2}-ax=0$.

Quiz problem:

A line meets the coordinate axes in A and B. A circle is circumscribed about the triangle OAB. If m and n are the distances of the tangent to the circle at the origin from the points A and B respectively, the diameter of the circle is (a) $m(m+n)$ (b) $m+n$ (c) $n(m+n)$ (d) $(1/2)(m+n)$.

To be continued,

Nalin Pithwa.

### Circles and Systems of Circles: IITJEE mains co-ordinate geometry basics

Section I:

Definition of a Circle:

A circle is the locus of  a point which moves in a plane so that it’s distance from a fixed point in the plane is always constant.The fixed point is called the centre of the circle and the constant distance is called its radius.

Section II:

Equations of a circle:

• An equation of a circle with centre $(h,k)$ and radius r is $(x-h)^{2}+(y-k)^{2}=r^{2}$.
• An equation of a circle with centre $(0,0)$ and radius r is $x^{2}+y^{2}=r^{2}$.
• An equation of a circle on the line segment joining $(x_{1},y_{1})$ and $(x_{2},y_{2})$ as diameter is $(x-x_{1})(x-x_{2})+(y-y_{1})(y-y_{2})=0$.
• General equation of a circle is :$x^{2}+y^{2}+2gx+2fy+c=0$ where g, f, and c are constants
• centre of this circle is $(-g,-f)$
• Its radius is $\sqrt{g^{2}+f^{2}-c}$, $g^{2}+f^{2}\geq c$
• Length of the intercept made by this circle on the x-axis is $2\sqrt{g^{2}-c}$ if $g^{2}-c \geq 0$ and that on the y-axis is $\sqrt{f^{2}-c}$ if $f^{2}-c \geq 0$.
• General equation of second order degree $ax^{2}+2hxy+by^{2}+2gx+2fy+c=0$ in x and y represents a circle if and only if:
• coefficient of $x^{2}$ equals coefficient of $y^{2}$, that is, $a=b \neq 0$
• coefficient of $xy$ is zero, that is , $h=0$
• $g^{2}+f^{2}-ac \geq 0$

Section III: Some results regarding circles:

• Position of a point with respect to a circle: Point $P(x_{1},y_{1})$ lies outside, on or inside a circle $S \equiv x^{2}+y^{2}+2gx+2fy+c=0$, according as $S_{1} \equiv x_{1}^{2}+y_{1}^{2}+2gx_{1}+2fy_{1}+c>, =, \hspace{0.1in} or \hspace{0.1in}< 0$
• Parametric coordinates of any point on the circle $(x-h)^{2}+(y-k)^{2}=r^{2}$ are given by $(h+r\cos{\theta},k+r\sin{\theta})$ with $0 \leq \theta < 2\pi$. In particular, parametric coordinates of any point on the circle.
• An equation of the tangent to the circle $x^{2}+y^{2}+2gx+2fy+c=0$ at the point $(x_{1},y_{1})$ on the circle is $xx_{1}+yy_{1}+g(x+x_{1})+f(y+y_{1})+c=0$
• An equation of the normal to the circle $x^{2}+y^{2}+2gx+2fy+c=0$ at the point $(x_{1},y_{1})$ on the circle is $\frac{y-y_{1}}{y_{1}+f} = \frac{x-x_{1}}{x_{1}+g}$
• Equations of the tangent and normal to the circle $x^{2}+y^{2}=r^{2}$ at the point $(x_{1},y_{1})$ on the circle are, respectively, $xx_{1}+yy_{1}=r^{2}$ and $\frac{x}{x_{1}}=\frac{y}{y_{1}}$
• The line $y=mx+c$ is a tangent to the circle $x^{2}+y^{2}=r^{2}$ if and only if $c^{2}=r^{2}(1+m^{2})$.
• The lines $y=mx \pm r\sqrt{(1+m^{2})}$ are tangents to the circle $x^{2}+y^{2}=r^{2}$, for all finite values of m. If m is infinite, the tangents are $x \pm r=0$.
• An equation of the chord of the circle $S=x^{2}+y^{2}+2gx+2fy+c=0$, whose mid-point is $(x_{1},y_{1})$ is $T=S_{1}$, where $T \equiv xx_{1}+yy_{1}+g(x+x_{1})+f(y+y_{1})+c$ and $S_{1}=x_{1}^{2}+y_{1}^{2}+2gx_{1}+2fy_{1}+c$. In particular, an equation of the chord of the circle $x^{2}+y^{2}=r^{2}$, whose mid-point is $(x_{1},y_{1})$ is $xx_{1}+yy_{1}=x_{1}^{2}+y_{1}^{2}$.
• An equation of the chord of contact of the tangents drawn from a point $(x_{1},y_{1})$ outside the circle $S=0$ is $T=0$.(S and T are as defined in (8) above).
• Length of the tangent drawn from a point $(x_{1},y_{1})$ outside the circle $S=0$, to the circle, is $\sqrt{S_{1}}$. (S and $\sqrt{S_{1}}$) are as defined in (8) above.)
• Two circles with centres $C_{1}(x_{1},y_{1})$ and $C_{2}(x_{2},y_{2})$ and radii $r_{1}$, $r_{2}$ respectively, (i) touch each other externally if $C_{1}C_{2}|=r_{1}+r_{2}$. the point of contact is $(\frac{r_{1}x_{2}+r_{2}x_{1}}{r_{1}+r_{2}},\frac{r_{1}y_{2}+r_{2}y_{1}}{r_{1}+r_{2}})$ and (ii) touch each other internally if $|C_{1}C_{2}|=|r_{1}-r_{2}|$, where $r_{1} \neq r_{2}$; the point of contact is $(\frac{r_{1}x_{2}-r_{2}x_{1}}{r_{1}-r_{2}} , \frac{r_{1}y_{2}-r_{2}y_{1}}{r_{1}-r_{2}})$
• An equation of the family of circles passing through the points $(x_{1},y_{1})$ and $(x_{2},y_{2})$ is $(x-x_{1})(x-x_{2}) + (y-y_{1})(y-y_{2}) +\lambda(F)=0$, where

$F=\left|\begin{array}{ccc} x & y & 1 \\ x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \end{array}\right|$

• An equation of the family of circles which touch the line $y-y_{1}=m(x-x_{1})$ at $(x_{1}, y_{1})$ for any finite value of m is $(x-x_{1})^{2}+(y-y_{1})^{2}+\lambda ((y-y_{1})-m(x-x_{1}))=0$. If m is infinite, the equation becomes $(x-x_{1})^{2}+(y-y_{1})^{2}+\lambda (x-x_{1})=0$.
• Let QR be a chord of a circle passing through the point $P(x_{1},y_{1})$ and let the tangents at the extremities Q and R of this chord intersect at the point $L(h,k)$. Then, locus of L is called the polar of P with respect to the circle, and P is called the pole of its polar.
• Equation of the polar of $P(x_{1},y_{1})$ with respect to the circle $S \equiv x^{2}+y^{2}+2gx+2fy+c=0$ is $T=0$, where T is defined as above.
• If the polar of P with respect to a circle passes through Q, then the polar of Q with respect to the same circle passes through P. Two such points P and Q, are called conjugate points of the same circle.
• If lengths of the tangents drawn from a point P to the two circles $S_{1} \equiv x^{2}+y^{2}+2g_{1}x+2f_{1}y+c_{1}=0$ and $S_{2} \equiv x^{2}+y^{2}+2g_{2}x+2f_{2}y+c_{2}=0$ are equal, then the locus of P is called the radical axis of the two circles $S_{1} =0$ and $S_{2}=0$, and its equation is $S_{1}-S_{2}=0$, that is, $2(g_{1}-g_{2})+2(f_{1}-f_{2})y+(c_{1}-c_{2})=0$
• Radical axis of two circles is perpendicular to the line joining their circles.
• Radical axes of three circles, taken in pairs, pass through a fixed point called the radical centre of the three circles, if the centres of these circles are non-collinear.

4: Special Forms of Equation of a Circle:

1. An equation of a circle with centre $(r,r)$ and radius $|r|$ is $(x-r)^{2}+(y-r)^{2}=r^{2}$. This touches the co-ordinate axes at the points $(r,0)$ and $(0,r)$.
2. An equation of a circle with centre $(x_{1},r)$, radius $|r|$ is $(x-x_{1})^{2}+(y-r)^{2}=r^{2}$. This touches the x-axis at $(x_{1},0)$.
3. An equation of a circle with centre $(\frac{a}{2},\frac{b}{2})$ and radius $\sqrt{\frac{(a^{2}+b^{2})}{4}}$ is $x^{2}+y^{2}-ax-by=0$. This circle passes through the origin $(0,0)$, and has intercepts a and b on the x and y axes, respectively.

5: Systems of Circles:

Let $S \equiv x^{2}+y^{2}+2gx+2fy+c$; and $S^{'} \equiv x^{2}+y^{2}+2g^{'}x+2f^{'}y+c^{'}$ and $L \equiv ax + by + k^{'}$.

1. If two circles $S=0$ and $S^{'}=0$ intersect at real and distinct points, then $S+\lambda S^{'}=0$ where $\lambda \neq -1$ represents a family of circles passing through these points, where $\lambda$ is a parameter, and $S-S^{'}=0$ when $\lambda=-1$ represents the chord of the circles.
2. If two circles $S =0$ and $S^{'}=0$ touch each other, then $S-S^{'}=0$ represents equation of the common tangent to the two circles at their point of contact.
3. If two circles $S=0$ and $S^{'}=0$ intersect each other orthogonally (the tangents at the point of intersection of the two circles are at right angles), then $2gg^{'}+2ff^{'}=c+c^{'}$.
4. If the circle $S=0$ intersects the line $L=0$ at two real and distinct points, then $S+\lambda L=0$ represents a family of circles passing through these points.
5. If $L=0$ is a tangent to the circle $S=0$ at P, then $S+\lambda L=0$ represents a family of circles touching $S=0$ at P, and having $L=0$ as the common tangent at P.
6. Coaxial Circles: A system of circles is said to be coaxial if every pair of circles of the system have the same radical axis. The simplest form of the equation of a coaxial system of circles is : $x^{2}+y^{2}+2gx+c=0$, where g is a variable and c is constant, the common radical axis of the system being y-axis and the line of centres being x-axis.  The Limiting points of the coaxial system of circles are the members of the system which are of zero radius. Thus, the limiting points of the coaxial system of circles $x^{2}+y^{2}+2gx+c=0$ are $(\pm \sqrt{c},0)$ if $c>0$. The equation $S+\lambda S^{'}=0$ ($\lambda \neq -1$) represents a family of coaxial circles, two of whose members are given to be $S=0$ and $S^{'}=0$.
7. Conjugate systems (or orthogonal systems) of circles : Two system of circles such that every circle of one system cuts every circle of  the other system orthogonally are said to  be conjugate system of circles. For instance, $x^{2}+y^{2}+2gx+c=0$ and $x^{2}+y^{2}+2fy-c=0$, where g and f are variables and c is constant, represent two systems of coaxial circles which are conjugate.

6: Common tangents to two circles:

If $(x-g_{1})^{2}+(y-f_{1})^{2}=a_{1}^{2}$ and $(x-g_{2})^{2}+(y-f_{2})^{2}=a_{2}^{2}$ are two circles with centres $C_{1}(g_{1},f_{1})$ and $C_{2}(g_{2},f_{2})$ and radii $a_{1}$ and $a_{2}$ respectively, then we have the following results regarding their common tangents:

1. When $C_{1}C_{2}>a_{1}+a_{2}$, that is, distance between the centres is greater than the sum of their radii, the two circles do not intersect with each other, and four common tangents can be drawn to circles. Two of them are direct common tangents and other two are transverse common tangents. The points $T_{1},T_{2}$ of intersection of direct common tangents and transverse common tangents respectively, always lie on the line joining the centres of the two circles and divide it externally and internally respectively in the ratio of their radii.
2. When $C_{1}C_{2}=a_{1}+a_{2}$, that is, the distance between the centres is equal to the sum of their radii, the two circles touch each other externally, two direct tangents are real and distinct and the transverse tangents coincide.
3. When $C_{1}C_{2}, that is, the distance between the centres is less than the sum of the radii, the circles intersect at two real and distinct points, the two direct common tangents are real and distinct while the transverse common tangents are imaginary.
4. When $C_{1}C_{2}=|a_{1}-a_{2}|$ with $a_{1} \neq a_{2}$, that is, the distance between the centres is equal to the difference of their radii, the circles touch each other internally, two direct common tangents are real and coincident, while the transeverse common tangents are imaginary.
5. When $C_{1}C_{2}<|a_{1}-a_{2}|$, with $a_{1} \neq a_{2}$, that is, the distance between the centres is less than the difference of the radii, one circle with smaller radius lies inside the other and the four common tangents are all imaginary.

To be continued,

Nalin Pithwa.

### Cartesian System, Straight Lines: IITJEE Mains: Problem Solving Skills II

I have a collection of some “random”, yet what I call ‘beautiful” questions in Co-ordinate Geometry. I hope kids preparing for IITJEE Mains or KVPY or ISI Entrance Examination will also like them.

Problem 1:

Given n straight lines and a fixed point O, a straight line is drawn through O meeting lines in the points $R_{1}$, $R_{2}$, $R_{3}$, $\ldots$, $R_{n}$ and on it a point R is taken such that $\frac{n}{OR} = \frac{1}{OR_{1}} + \frac{1}{OR_{2}} + \frac{1}{OR_{3}} + \ldots + \frac{1}{OR_{n}}$

Show that the locus of R is a straight line.

Solution 1:

Let equations of the given lines be $a_{i}x+b_{i}y+c_{i}=0$, $i=1,2,\ldots, n$, and the point O be the origin $(0,0)$.

Then, the equation of the line through O can be written as $\frac{x}{\cos{\theta}} = \frac{y}{\sin{\theta}} = r$ where $\theta$ is the angle made by the line with the positive direction of x-axis and r is the distance of any point on the line from the origin O.

Let $r, r_{1}, r_{2}, \ldots, r_{n}$ be the distances of the points $R, R_{1}, R_{2}, \ldots, R_{n}$ from O which in turn $\Longrightarrow OR=r$ and $OR_{i}=r_{i}$, where $i=1,2,3 \ldots n$.

Then, coordinates of R are $(r\cos{\theta}, r\sin{\theta})$ and of $R_{i}$ are $(r_{i}\cos{\theta},r_{i}\sin{\theta})$ where $i=1,2,3, \ldots, n$.

Since $R_{i}$ lies on $a_{i}x+b_{i}y+c_{i}=0$, we can say $a_{i}r_{i}\cos{\theta}+b_{i}r_{i}\sin{\theta}+c_{i}=0$ for $i=1,2,3, \ldots, n$

$\Longrightarrow -\frac{a_{i}}{c_{i}}\cos{\theta} - \frac{b_{i}}{c_{i}}\sin{\theta} = \frac{1}{r_{i}}$, for $i=1,2,3, \ldots, n$

$\Longrightarrow \sum_{i=1}^{n}\frac{1}{r_{i}}=-(\sum_{i=1}^{n}\frac{a_{i}}{c_{i}})\cos{\theta}-(\sum_{i=1}^{n}\frac{b_{i}}{c_{i}})\sin{\theta}$

$\frac{n}{r}=-(\sum_{i=1}^{n}\frac{a_{i}}{c_{i}})\cos{\theta}-(\sum_{i=1}^{n}\frac{b_{i}}{c_{i}})\sin{\theta}$ …as given…

$\Longrightarrow (\sum_{i=1}^{n}\frac{a_{i}}{c_{i}})r\cos{\theta}+(\sum_{i=1}^{n}\frac{b_{i}}{c_{i}})r\sin{\theta} + n=0$

Hence, the locus of R is $(\sum_{i=1}^{n}\frac{a_{i}}{c_{i}})x+(\sum_{i=1}^{n}\frac{b_{i}}{c_{i}})y+n=0$ which is a straight line.

Problem 2:

Determine all values of $\alpha$ for which the point $(\alpha,\alpha^{2})$ lies inside the triangle formed by the lines $2x+3y-1=0$, $x+2y-3=0$, $5x-6y-1=0$.

Solution 2:

Solving equations of the lines two at a time, we get the vertices of the given triangle as: $A(-7,5)$, $B(1/3,1/9)$ and $C(5/4, 7/8)$.

So, AB is the line $2x+3y-1=0$, AC is the line $x+2y-3=0$ and BC is the line $5x-6y-1=0$

Let $P(\alpha,\alpha^{2})$ be a point inside the triangle ABC. (please do draw it on a sheet of paper, if u want to understand this solution further.) Since A and P lie on the same side of the line $5x-6y-1=0$, both $5(-7)-6(5)-1$ and $5\alpha-6\alpha^{2}-1$ must have the same sign.

$\Longrightarrow 5\alpha-6\alpha^{2}-1<0$ or $6\alpha^{2}-5\alpha+1>0$ which in turn $\Longrightarrow (3\alpha-1)(2\alpha-1)>0$ which in turn $\Longrightarrow$ either $\alpha<1/3$ or $\alpha>1/2$….call this relation I.

Again, since B and P lie on the same side of the line $x+2y-3=0$, $(1/3)+(2/9)-3$ and $\alpha+2\alpha^{2}-3$ have the same sign.

$\Longrightarrow 2\alpha^{2}+\alpha-3<0$ and $\Longrightarrow (2\alpha+3)(\alpha-1)<0$, that is, $-3/2 <\alpha <1$…call this relation II.

Lastly, since C and P lie on the same side of the line $2x+3y-1=0$, we have $2 \times (5/4) + 3 \times (7/8) -1$ and $2\alpha+3\alpha^{2}-1$ have the same sign.

$\Longrightarrow 3\alpha^{2}+2\alpha-1>0$ that is $(3\alpha-1)(\alpha+1)>0$

$\alpha<-1$ or $\alpha>1/3$….call this relation III.

Now, relations I, II and III hold simultaneously if $-3/2 < \alpha <-1$ or $1/2<\alpha<1$.

Problem 3:

A variable straight line of slope 4 intersects the hyperbola $xy=1$ at two points. Find the locus of the point which divides the line segment between these two points in the ratio $1:2$.

Solution 3:

Let equation of the line be $y=4x+c$ where c is a parameter. It intersects the hyperbola $xy=1$ at two points, for which $x(4x+c)=1$, that is, $\Longrightarrow 4x^{2}+cx-1=0$.

Let $x_{1}$ and $x_{2}$ be the roots of the equation. Then, $x_{1}+x_{2}=-c/4$ and $x_{1}x_{2}=-1/4$. If A and B are the points of intersection of the line and the hyperbola, then the coordinates of A are $(x_{1}, \frac{1}{x_{1}})$ and that of B are $(x_{2}, \frac{1}{x_{2}})$.

Let $R(h,k)$ be the point which divides AB in the ratio $1:2$, then $h=\frac{2x_{1}+x_{2}}{3}$ and $k=\frac{\frac{2}{x_{1}}+\frac{1}{x_{2}}}{3}=\frac{2x_{2}+x_{1}}{3x_{1}x_{2}}$, that is, $\Longrightarrow 2x_{1}+x_{2}=3h$…call this equation I.

and $x_{1}+2x_{2}=3(-\frac{1}{4})k=(-\frac{3}{4})k$….call this equation II.

Adding I and II, we get $3(x_{1}+x_{2})=3(h-\frac{k}{4})$, that is,

$3(-\frac{c}{4})=3(h-\frac{k}{4}) \Longrightarrow (h-\frac{k}{4})=-\frac{c}{4}$….call this equation III.

Subtracting II from I, we get $x_{1}-x_{2}=3(h+\frac{k}{4})$

$\Longrightarrow (x_{1}-x_{2})^{2}=9(h+\frac{k}{4})^{2}$

$\Longrightarrow \frac{c^{2}}{16} + 1= 9(h+\frac{k}{4})^{2}$

$\Longrightarrow (h-\frac{k}{4})^{2}+1=9(h+\frac{k}{4})^{2}$

$\Longrightarrow h^{2}-\frac{1}{2}hk+\frac{k^{2}}{16}+1=9(h^{2}+\frac{1}{2}hk+\frac{k^{2}}{16})$

$\Longrightarrow 16h^{2}+10hk+k^{2}-2=0$

so that the locus of $R(h,k)$ is $16x^{2}+10xy+y^{2}-2=0$

More later,

Nalin Pithwa.

### Cartesian system and straight lines: IITJEE Mains: Problem solving skills

Problem 1:

The line joining $A(b\cos{\alpha},b\sin{\alpha})$ and $B(a\cos{\beta},a\sin{\beta})$ is produced to the point $M(x,y)$ so that $AM:MB=b:a$, then find the value of $x\cos{\frac{\alpha+\beta}{2}}+y\sin{\frac{\alpha+\beta}{2}}$.

Solution 1:

As M divides AB externally in the ratio $b:a$, we have $x=\frac{b(a\cos{\beta})-a(b\cos{\alpha})}{b-a}$ and $y=\frac{b(a\sin{\beta})-a(b\sin{\alpha})}{b-a}$ which in turn

$\Longrightarrow \frac{x}{y} = \frac{\cos{\beta}-cos{\alpha}}{\sin{\beta}-\sin{\alpha}}$

$= \frac{2\sin{\frac{\alpha+\beta}{2}}\sin{\frac{\alpha-\beta}{2}}}{2\cos{\frac{\alpha+\beta}{2}}\sin{\frac{\beta-\alpha}{2}}}$

$\Longrightarrow x\cos{\frac{\alpha+\beta}{2}}+y\sin{\frac{\alpha+\beta}{2}}=0$

Problem 2:

If the circumcentre of a triangle lies at the origin and the centroid in the middle point of the line joining the points $(a^{2}+1,a^{2}+1)$ and $(2a,-2a)$, then where does the orthocentre lie?

Solution 2:

From plane geometry, we know that the circumcentre, centroid and orthocentre of a triangle lie on a line. So, the orthocentre of the triangle lies on the line joining the circumcentre $(0,0)$ and the centroid $(\frac{(a+1)^{2}}{2},\frac{(a-1)^{2}}{2})$, that is, $y.\frac{(a+1)^{2}}{2} = x.\frac{(a-1)^{2}}{2}$, or $(a-1)^{2}x-(a+1)^{2}y=0$. That is, the orthocentre lies on this line.

Problem 3:

If a, b, c are unequal and different from 1 such that the points $(\frac{a^{3}}{a-1},\frac{a^{2}-3}{a-1})$, $(\frac{b^{3}}{b-1},\frac{b^{2}-3}{b-1})$ and $(\frac{c^{3}}{c-1},\frac{c^{2}-3}{c-1})$ are collinear, then which of the following option is true?

a: $bc+ca+ab+abc=0$

b: $a+b+c=abc$

c: $bc+ca+ab=abc$

d: $bc+ca+ab-abc=3(a+b+c)$

Solution 3:

Suppose the given points lie on the line $lx+my+n=0$ then a, b, c are the roots of the equation :

$lt^{3}+m(t^{2}-3)+n(t-1)=0$, or

$lt^{3}+mt^{2}+nt-(3m+n)=0$

$\Longrightarrow a+b+c=-\frac{m}{l}$ and $ab+bc+ca=\frac{n}{l}$, that is, $abc=(3m+n)/l$

Eliminating l, m, n, we get $abc=-3(a+b+c)+bc+ca+ab$

$\Longrightarrow bc+ca+ab-abc=3(a+b+c)$, that is, option (d) is the answer.

Problem 4:

If $p, x_{1}, x_{2}, \ldots, x_{i}, \ldots$ and $q, y_{1}, y_{2}, \ldots, y_{i}, \ldots$ are in A.P., with common difference a and b respectively, then on which line does the centre of mean position of the points $A_{i}(x_{i},y_{i})$ with $i=1,2,3 \ldots, n$ lie?

Solution 4:

Note: Centre of Mean Position is $(\frac{\sum{xi}}{n},\frac{\sum {yi}}{n})$.

Let the coordinates of the centre of mean position of the points $A_{i}$, $i=1,2,3, \ldots,n$ be $(x,y)$ then

$x=\frac{x_{1}+x_{2}+x_{3}+\ldots + x_{n}}{n}$ and $y=\frac{y_{1}+y_{2}+\ldots + y_{n}}{n}$

$\Longrightarrow x = \frac{np+a(1+2+\ldots+n)}{n}$, $y=\frac{nq+b(1+2+\ldots+n)}{n}$

$\Longrightarrow x=p+ \frac{n(n+1)}{2n}a$ and $y=q+ \frac{n(n+1)}{2n}b$

$\Longrightarrow x=p+\frac{n+1}{2}a$, and $y=q+\frac{n+1}{2}b$

$\Longrightarrow 2\frac{(x-p)}{a}=2\frac{(y-q)}{b} \Longrightarrow bx-ay=bp-aq$, that is, the CM lies on this line.

Problem 5:

The line L has intercepts a and b on the coordinate axes. The coordinate axes are rotated through a fixed angle, keeping the origin fixed. If p and q are the intercepts of the line L on the new axes, then what is the value of $\frac{1}{a^{2}} - \frac{1}{p^{2}} + \frac{1}{b^{2}} - \frac{1}{q^{2}}$?

Solution 5:

Equation of the line L in the two coordinate systems is $\frac{x}{a} + \frac{y}{b}=1$, and $\frac{X}{p} + \frac{Y}{q}=1$ where $(X,Y)$ are the new coordinate of a point $(x,y)$ when the axes are rotated through a fixed angle, keeping the origin fixed. As the length of the perpendicular from the origin has not changed.

$\frac{1}{\sqrt{\frac{1}{a^{2}}+\frac{1}{b^{2}}}}=\frac{1}{\sqrt{\frac{1}{p^{2}} + \frac{1}{q^{2}}}}$

$\Longrightarrow \frac{1}{a^{2}} + \frac{1}{b^{2}} = \frac{1}{p^{2}} + \frac{1}{q^{2}}$

or $\frac{1}{a^{2}} - \frac{1}{p^{2}} + \frac{1}{b^{2}} - \frac{1}{q^{2}}=0$. So, the value is zero.

Problem 6:

Let O be the origin, $A(1,0)$ and $B(0,1)$ and $P(x,y)$ are points such that $xy>0$ and $x+y<1$, then which of the following options is true:

a: P lies either inside the triangle OAB or in the third quadrant

b: P cannot lie inside the triangle OAB

c: P lies inside the triangle OAB

d: P lies in the first quadrant only.

Solution 6:

Since $xy>0$, P either lies in the first quadrant or in the third quadrant. The inequality $x+y<1$ represents all points below the line $x+y=1$. So that $xy>0$ and $x+y<1$ imply that either P lies inside the triangle OAB or in the third quadrant.

Problem 7:

An equation of a line through the point $(1,2)$ whose distance from the point $A(3,1)$ has the greatest value is :

option i: $y=2x$

option ii: $y=x+1$

option iii: $x+2y=5$

option iv: $y=3x-1$

Solution 7:

Let the equation of the line through $(1,2)$ be $y-2=m(x-1)$. If p denotes the length of the perpendicular from $(3,1)$ on this line, then $p=|\frac{2m+1}{\sqrt{m^{2}+1}}|$

$\Longrightarrow p^{2}=\sqrt{\frac{4m^{2}+4m+1}{m^{2}+1}}=4+ \frac{4m-3}{m^{2}+1}=s$, say

then $p^{2}$ is greatest if and only if s is greatest.

Now, $\frac{ds}{dm} = \frac{(m^{2}+1)(4)-2m(4m-3)}{(m^{2}+1)^{2}} = \frac{-2(2m-1)(m-2)}{(m^{2}+1)^{2}}$

$\frac{ds}{dm} = 0$ so that $\Longrightarrow m = \frac{1}{2}, 2$. Also, $\frac{ds}{dm}<0$, if $m<\frac{1}{2}$, and

$\frac{ds}{dm} >0$, if $1/2

and $\frac{ds}{dm} <0$, if $m>2$. So s is greatest for $m=2$. And, thus, the equation of the required line is $y=2x$.

Problem 8:

The points $A(-4,-1)$, $B(-2,-4)$, Slatex C(4,0)\$ and $D(2,3)$ are the vertices of a :

option a: parallelogram

option b: rectangle

option c: rhombus

option d: square.

Note: more than one option may be right. Please mark all that are right.

Solution 8:

Mid-point of AC = $(\frac{-4+4}{2},\frac{-1+0}{2})=(0, \frac{-1}{2})$

Mid-point of BD = $(\frac{-2+2}{2},\frac{-4+3}{2})=(0,\frac{-1}{2})$

$\Longrightarrow$ the diagonals AC and BD bisect each other.

$\Longrightarrow$ ABCD is a parallelogram.

Next, $AC= \sqrt{(-4-4)^{2}+(-1+0)^{2}}=\sqrt{64+1}=\sqrt{65}$ and $BD=\sqrt{(-2-2)^{2}+(-4+3)^{2}}=\sqrt{16+49}=\sqrt{65}$ and since the diagonals are also equal, it is a rectangle.

As $AB=\sqrt{(-4+2)^{2}+(-1+4)^{2}}=\sqrt{13}$ and $BC=\sqrt{(-2-4)^{2}+(-4)^{2}}=\sqrt{36+16}=sqrt{52}$, the adjacent sides are not equal and hence, it is neither a rhombus nor a square.

Problem 9:

Equations $(b-c)x+(c-a)y+(a-b)=0$ and $(b^{3}-c^{3})x+(c^{3}-a^{3})y+a^{3}-b^{3}=0$ will represent the same line if

option i: $b=c$

option ii: $c=a$

option iii: $a=b$

option iv: $a+b+c=0$

Solution 9:

The two lines will be identical if there exists some real number k, such that

$b^{3}-c^{3}=k(b-c)$, and $c^{3}-a^{3}=k(c-a)$, and $a^{3}-b^{3}=k(a-b)$.

$\Longrightarrow b-c=0$ or $b^{2}+c^{2}+bc=k$

$\Longrightarrow c-a=0$ or $c^{2}+a^{2}+ac=k$, and

$\Longrightarrow a-b=0$ or $a^{2}+b^{2}+ab=k$

That is, $b=c$ or $c=a$, or $a=b$.

Next, $b^{2}+c^{2}+bc=c^{2}+a^{2}+ca \Longrightarrow b^{2}-a^{2}=c(a-b)$. Hence, $a=b$, or $a+b+c=0$.

Problem 10:

The circumcentre of a triangle with vertices $A(a,a\tan{\alpha})$, $B(b, b\tan{\beta})$ and $C(c, c\tan{\gamma})$ lies at the origin, where $0<\alpha, \beta, \gamma < \frac{\pi}{2}$ and $\alpha + \beta + \gamma = \pi$. Show that it’s orthocentre lies on the line $4\cos{\frac{\alpha}{2}}\cos{\frac{\beta}{2}}\cos{\frac{\gamma}{2}}x-4\sin{\frac{\alpha}{2}}\sin{\frac{\beta}{2}}\sin{\frac{\gamma}{2}}y=y$

Solution 10:

As the circumcentre of the triangle is at the origin O, we have $OA=OB=OC=r$, where r is the radius of the circumcircle.

Hence, $OA^{2}=r^{2} \Longrightarrow a^{2}+a^{2}\tan^{2}{\alpha}=r^{2} \Longrightarrow a = r\cos{\alpha}$

Therefore, the coordinates of A are $(r\cos{\alpha},r\sin{\alpha})$. Similarly, the coordinates of B are $(r\cos{\beta},r\sin{\beta})$ and those of C are $(r\cos{\gamma},r\sin{\gamma})$. Thus, the coordinates of the centroid G of $\triangle ABC$ are

$(\frac{1}{3}r(\cos{\alpha}+\cos{\beta}+\cos{\gamma}),\frac{1}{3}r(\sin{\alpha}+\sin{\beta}+\sin{\gamma}))$.

Now, if $P(h,k)$ is the orthocentre of $\triangle ABC$, then from geometry, the circumcentre, centroid, and the orthocentre of a triangle lie on a line, and the slope of OG equals the slope of OP.

Hence, $\frac{\sin{\alpha}+\sin{\beta}+\sin{\gamma}}{\cos{\alpha}+\cos{\beta}+\cos{\gamma}}=\frac{k}{h}$

$\Longrightarrow \frac{4\cos{(\frac{\alpha}{2})}\cos{(\frac{\beta}{2})}\cos{(\frac{\gamma}{2})}}{1+4\sin{(\frac{\alpha}{2})}\sin{(\frac{\beta}{2})}\sin{(\frac{\gamma}{2})}}= \frac{k}{h}$

because $\alpha+\beta+\gamma=\pi$.

Hence, the orthocentre $P(h,k)$ lies on the line

$4\cos{(\frac{\alpha}{2})}\cos{(\frac{\beta}{2})}\cos{(\frac{\gamma}{2})}x-4\sin{(\frac{}{})}\sin{(\frac{\beta}{2})}\sin{(\frac{\gamma}{2})}y=y$.

Hope this gives an assorted flavour. More stuff later,

Nalin Pithwa.