The length of tangent from point $(5, 1)$ to the circle $x^{2} + y^{2} + 6 x - 4 y - 3 = 0$ is:

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Solution

The correct option is B 7
Given equation of circle $x^{2} + y^{2} + 6 x - 4 y - 3 = 0$
Length of tangent $= \sqrt{S_{1}}$
$= \sqrt{5^{2} + 1^{2} + 30 - 4 - 3}$
$= \sqrt{49} = 7$

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(5, 1)

x^{2} + y^{2} + 6 x - 4 y - 3 = 0

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(2, 2)

x^{2} + y^{2} - 6 x - 4 y + 3 = 0

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x^{2} + y^{2} - 6 x - 4 y - 11 = 0

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