Question

The polar of the point on the circle $x^2+y^2=p^2$ with respect to the circle $x^2+y^2=q^2$ touches the circle $x^2+y^2=r^2$, their $p,q,r$ are in$\cdots \cdots$ progression

• A. Arithmetic
• B. Geometric
• C. Harmonic
• D. Arithmetic, Geometric

Answer 1

Answer: (B)

Geometric

Let polar of the point $P(p\cos \theta ,p\sin \theta )$ on $x^2+y^2=q^2$ is
$xp\cos \theta +yp\sin \theta =q^2$ ----(1)
and $xp\cos \theta +yp\sin \theta =q^2=0$ is a tangent to a $x^2+y^2=r^2$
$\therefore r=\left|\frac{-q^2}{\sqrt{p^2{\cos }^2\theta +p^2{\sin }^2\theta }}\right|$
$rp=q^2$
$\Rightarrow q=\sqrt{rp}$
$p,q,r$ are in geometric progression