Question

The distances from the origin of the centers of three circles $x^2+y^2-2\lambda x=c^2$ (where $c$ is a constant and $\lambda$ a variable) are in geometrical progression; prove that the lengths of the tangents drawn to them from any point on the circle $x^2+y^2=c^2$ are also in geometrical progression.

Answer 1

Given circles are

$x^2+y^2-2{\lambda }_{1}x-c^2=\mathrm{0......}\left(i\right)x^2+y^2-2{\lambda }_{2}x-c^2=\mathrm{0.......}\left(ii\right)x^2+y^2-2{\lambda }_{3}x-c^2=\mathrm{0........}\left(iii\right)$

Distance of their centres from origin is ${\lambda }_1,{\lambda }_2$ and ${\lambda }_3$ respectively

Given ${\lambda }_2^2={\lambda }_1{\lambda }_3$

let tangents are drawn from $(h,k)$ on $x^2+y^2-c^2=0$ to all the three circles

$h^2+k^2-c^2=\mathrm{0........}\left(iv\right)$

Length of tangent from $(h,k)$ to $\left(i\right)$ is

${PT}_1=\sqrt{h^2+k^2-2{\lambda }_{1}h-c^2}{PT}_1=\sqrt{h^2+k^2-c^2-2{\lambda }_{1}h}$

Substituting $\left(iv\right)$ we get

${PT}_1=\sqrt{0-2{\lambda }_{1}h}=\sqrt{-2{\lambda }_{1}h}$

$\Rightarrow {PT}_1^2=-2{\lambda }_{1}h$

Similarly ${PT}_2^2=-2{\lambda }_{2}h$ and ${PT}_3^2=-2{\lambda }_{3}h$

If lengths of tangents are in G.P. then square of their lengths will also be in G.P.

$\Rightarrow {PT}_2^4={PT}_1^2\times {PT}_3^2$

Substituting the values

$\Rightarrow {(-2{\lambda }_{2}h)}^2=-2{\lambda }_{1}h\times -2{\lambda }_{3}h\Rightarrow {\lambda }_2^2={\lambda }_1{\lambda }_3$

which is required.

Hence proved that their lengths are in G.P.