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(B) The equation of the circles is $x^2 + y^2 - 2\lambda_i x - c^2 = 0$.The center of each circle is $(\lambda_i, 0)$ and the distance from the origin

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$G.P.$

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(B) The equation of the circles is $x^2 + y^2 - 2\lambda_i x - c^2 = 0$.The center of each circle is $(\lambda_i, 0)$ and the distance from the origin is $|\lambda_i|$.Since $|\lambda_1|, |\lambda_2|, |\lambda_3|$ are in $G.P.$,we have $\lambda_2^2 = \lambda_1 \lambda_3$.Let $(x_0, y_0)$ be any point on the circle $x^2 + y^2 = c^2$,so $x_0^2 + y_0^2 = c^2$.The length of the tangent $L_i$ from $(x_0, y_0)$ to the circle $x^2 + y^2 - 2\lambda_i x - c^2 = 0$ is given by $L_i = \sqrt{x_0^2 + y_0^2 - 2\lambda_i x_0 - c^2}$.Substituting $x_0^2 + y_0^2 = c^2$,we get $L_i = \sqrt{c^2 - 2\lambda_i x_0 - c^2} = \sqrt{-2\lambda_i x_0}$.Since $L_i^2 = -2\lambda_i x_0$,the squares of the lengths of the tangents are proportional to $\lambda_i$.Since $\lambda_i$ are in $G.P.$,their squares $L_i^2$ are also in $G.P.$,which implies the lengths $L_i$ are in $G.P.$

List-$I$	List-$II$
$(i)$ The equation of the polar of $(-5, 1)$ with respect to $C$	$(A)$ $y = 0$
$(ii)$ The equation of the tangent at $(8, 0)$ to $C$	$(B)$ $y = 6$
$(iii)$ The equation of the normal at $(2, 6)$ to $C$	$(C)$ $x + y = 7$
$(iv)$ The equation of the diameter of $C$ through $(8, 12)$	$(D)$ $13x + 5y = 98$
	$(E)$ $x = 8$

If the distances from the origin to the centres of the three circles $x^2 + y^2 - 2\lambda_i x = c^2$ for $i = 1, 2, 3$ 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

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