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

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$A . G . P$

$G . P$

$A . P$

$H . P$

answer is B.

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Detailed Solution

$x^{2} + y^{2} + 2 λ_{i} x - c^{2} = 0 {λ_{i}}^{2} + c^{2} Centre is (- λ_{i}, 0) Radius is \sqrt{{λ_{i}}^{2} + c^{2}} Distance between origin and centre = \sqrt{{λ_{i}}^{2}} = λ_{i} Distances are in G . P . \Rightarrow {λ_{2}}^{2} = λ_{1} λ_{3} \Rightarrow \frac{λ_{2}}{λ_{1}} = \frac{λ_{3}}{λ_{2}} \dots \dots (1) Any point on x^{2} + y^{2} = c^{2} is P (ccosθ, csinθ) Length of tangent to C_{1} is PA = \sqrt{S_{11}} = \sqrt{2 λ_{1} cosθ} Similarly, length of tangent to circle C_{2} is PB = \sqrt{2 λ_{2} cosθ} Similarly, length of tangent to circle C_{3} is PC = \sqrt{2 λ_{3} cosθ} Now, \frac{PB}{PA} = \sqrt{\frac{λ_{2}}{λ_{1}}} \frac{PC}{PB} = \sqrt{\frac{λ_{3}}{λ_{1}}} From equation (1), we get \frac{PB}{PA} = \frac{PC}{PB} \Rightarrow {(PB)}^{2} = PA \cdot PC Therefore, PA, PB, PC are in G . P$