Question:medium

The point $ (3, -4) $ lies on both the circles $ x^{2} + y^{2} - 2x + 8y + 13 = 0 $ and $ x^{2} + y^{2} - 4x + 6y + 11 = 0 $. Then, the angle between the circles is

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Use $\cos \theta = \frac{d^{2}-r_{1}^{2}-r_{2}^{2}}{2r_{1}r_{2}}$ to find the angle between two intersecting circles.

Updated On: Apr 10, 2026

$60^{\circ}$
$\tan^{-1}(\frac{1}{2})$
$\tan^{-1}(\frac{3}{5})$
$135^{\circ}$

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The Correct Option is D

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The point \( (3, -4) \) lies on both the circles \( x^{2} + y^{2} - 2x + 8y + 13 = 0 \) and \( x^{2} + y^{2} - 4x + 6y + 11 = 0 \). Then, the angle between the circles is

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The Correct Option is D

Solution and Explanation

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