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) $x-y=0$

Answer 1: C.

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) $x+2y=0$ .

Answer 2: C.

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) $x^{2}+ y^{2}-ay=0$ .

Answer c.

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.

Mathematics Hothouse