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If the distances from the origin to 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 the tangents drawn to them from any point on the circle $x^2+y^2=c^2$ are in
Option: 1
A.P.

Option: 2
G.P.

Option: 3
H.P.

Option: 4
None of these

Answers (1)

best_answer

The centres of the given circles are $\left(-\lambda_i, 0\right)(i=1,2,3)$

The distances from the origin to the centres are $\lambda_i(i=1,2,3)$ . It is given that $\lambda_2^2=\lambda_1 \lambda_3 .$

Let $P(h, k)$ be any point on the circle $x^2+y^2=c^2$ ,then $h^2+k^2=c^2$

Now, $L_i=$ length of the tangent from $(h, k)$ to $x^2+y^2+2 \lambda_i x-c^2=0$

$=\sqrt{h^2+k^2+2 \lambda_i h-c^2}=\sqrt{c^2+2 \lambda_i h-c^2}=\sqrt{2 \lambda_i h}$ $[ h^2+k^2=c^2 \text { and } i=1,2,3 \text { ] }$

Therefore, $L_2^2=2 \lambda_2 h=2 h\left(\sqrt{\lambda_1 \lambda_3}\right)$ $\text {[ } \quad \lambda_2^2=\lambda_1 \lambda_3 \text { ] }$

$=\sqrt{2 h \lambda_1} \sqrt{2 h \lambda_3}=L_1 L_3 \text {. Hence, } L_1, L_2, L_3 \text { are in G.P. }$

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