Question

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

• A. A.P.
• B. G.P
• C. H.P
• D. A.G.P
  

Answer 1

The correct option is B G.P

$x^2+y^2-2kix=c^2$
$centre,C\equiv (k_i,0)$
so, distance of centre of circle C from origin $(0,0)$ is
$d=\sqrt{k{i}^2}=k_i$
$\therefore$ as given distance from the origin to the centres of three circle
$x^2+y^2-2k_{i}x=C^2$ are in G.P
then $k_1,k_2,k_3$ are in G.P
$k_2^2=k_1{k}_3-\left(1\right)$
let ,$P(\cos ,c\sin )$ be any point on $x^2+y^2=C^2$ circle.
Then length of tangents to three circles from P is
$L=\sqrt{5_1}=\sqrt{c^2-c^2}-2k_1\sin \times C$
$=\sqrt{-2k_{1}c\sin }$
$\therefore L1=\sqrt{-2k_{1}C\sin }$
$L2=\sqrt{-2k_{2}C\sin }$
$L_3=\sqrt{-2k_{3}C\sin }$
Then $L_1{L}_2$ and $L_3$ are also in G.P
Because $k_1,k_2,k_3$ are in G.P
Then $\sqrt{k_1}\sqrt{k_2}\sqrt{k_3}$are also in G.P