Question

Distance from the origin to the centres of the three circles $x^2+y^2-2\lambda x=c^2$ (where $c$ is constant and $\lambda$ is variable) are in $G.P.$ Prove that the lengths of tangents drawn from any point on the circle $x^2+y^2=c^2$ to the three circles are also in $G.P.$.

Answer 1

If ${\lambda }_1,{\lambda }_2,{\lambda }_3$ be the three values of $\lambda$, then
${\lambda }_2^2={\lambda }_1{\lambda }_3....\left(1\right)$
If $(h,k)$, be any point on the circle $x^2+y^2=c^2$
then $h^2+k^2=c^2.....\left(2\right)$
Suppose $t_1,t_2,t_3$ be the lengths of tangents from $(h,k)$ to the circle
$x^2+y^2-2{\lambda }_{r}x-c^2=0,(r=1,2,3)$ then
$t_1^2=h^2+k^2-2{\lambda }_{1}h-c^2=-2{\lambda }_{1}h$ by $\left(2\right)$.
Similarly $t_2^2=-2{\lambda }_{2}h,t_3^2=-2{\lambda }_{3}h$.
Now $t_1,t_2,t_3$ will be in G.P. if $t_1^2,t_2^2$ and $t_3^2$ are also in G.P.
or $(-2{\lambda }_{2}h{)}^2=(-2{\lambda }_{1}h\left)\right(-2{\lambda }_{3}h)$.
or ${\lambda }_2^2={\lambda }_1{\lambda }_3$ which is true by $\left(1\right)$.