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Number of Common Tangents to Two Circles in Different Conditions

The point fro...

Question

The point from which the tangents to the circles $x^{2} + y^{2} - 8 x + 40 = 0, 5 x^{2} + 5 y^{2} - 25 x + 80 = 0, x^{2} + y^{2} - 8 x + 16 y + 160 = 0$ are equal in length is

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Solution

The correct option is C

Required point is the radical centre of the three given circles. The radical axes of these three circles taken in pairs are

3x – 24 = 0, 16y + 120 = 0, 3x + 16y + 80 = 0

∴ x = 8, y = $\frac{- 15}{2}$

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x^{2} + y^{2} = 1, x^{2} + y^{2} + 8 x + 15 = 0, x^{2} + y^{2} + 10 y + 24 = 0

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I . x^{2} + y^{2} = 1

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