Question

A tangent at a point on the circle $x^2+y^2=a^2$ intersect a concentric circle S at P and Q. The tangents of this circle at, P,Q meet on the circle $x^2+y^2=b^2$. The equation of the concentric circle S is

• A. $x^2+y^2=a^2+b^2$
• B. $x^2+y^2=a^2{b}^2$
• C. $x^2+y^2=a^2-b^2$
• D. $x^2+y^2=ab$

Answer 1

Answer: (D)

$x^2+y^2=ab$

Take, any point on $x^2+y^2=b^2$ as P($bcos\theta ,bsin\theta )$

From P $(bcos\theta ,bsin\theta )$ take
a chord of contact of circle S.

This chord of contact is tangent to $x^2+y^2=a^2$

Chord of contact to circle S: $x^2+y^2=r^2$

is

$x{x}_1+y{y}_1-r^2=0$

$xbcos\theta +ybsin\theta -r^2=0$

$\therefore$ distance of (0,0) to $xbcos\theta+b

sin \theta y-r^{2}=0$ is a.

$|\frac{r^2}{\sqrt{b^2}}|=a$

$r^2=ab.$

$\therefore r=\sqrt{ab}$

$\therefore$ equation of circle S is

$x^2+y^2=r^2$

$\to x^2+y^2=ab$