Question

If the distances from the origin of the centres of three circles $x^2+y^2+2{\lambda }_{i}x-c^2=0$ $(i=1,2,3)$ are in G.P., then the lengths of tangents drawn to them from any point on the circle $x^2+y^2=c^2$ are in

• A. A.P.
• B. G.P.
• C. H.P.
• D. None of these

Answer 1

The correct option is B G.P.

$x^2+y^2+2{\lambda }_{1}x-c^2=0$
Centre is $(-{\lambda }_i,0)$
Radius is $\sqrt{{\lambda }_i^2+c^2}$
Distance between origin and centre
$=\sqrt{(-{\lambda }_i{)}^2+0}={\lambda }_i$
Distances are in G.P.
$\Rightarrow {\lambda }_2^2={\lambda }_1{\lambda }_3$
$\Rightarrow \frac{{\lambda }_2}{{\lambda }_1}=\frac{{\lambda }_3}{{\lambda }_2}\dots \dots \left(1\right)$

Any point on $x^2+y^2=c^2$ is $P(c\cos \theta ,c\sin \theta )$
Length of tangent to $C_1$ is
$PA=\sqrt{S_1}=\sqrt{2{\lambda }_{1}c\cos \theta }$
Similarly, length of tangent to circle $C_2$ is
$PB=\sqrt{2{\lambda }_{2}c\cos \theta }$
Similarly, length of tangent to circle $C_3$ is
$PC=\sqrt{2{\lambda }_{3}c\cos \theta }$

Now,
$\frac{PB}{PA}=\sqrt{\frac{{\lambda }_2}{{\lambda }_1}}$
$\frac{PC}{PB}=\sqrt{\frac{{\lambda }_3}{{\lambda }_2}}$

From equation $\left(1\right)$, we get
$\frac{PB}{PA}=\frac{PC}{PB}\Rightarrow (PB{)}^2=PA\cdot PC$

Therefore, $PA,PB,PC$ are in G.P.