## Tag Archives: descartes

### 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.