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

7

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49

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63

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21

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Solution

The correct option is A $7$
$S_{1} : x^{2} + y^{2} + 6 x - 4 y - 3 = 0$
Center of the above circle, $S_{1}$ , is $C (- 3, 2)$ and radius $r = \sqrt{9 + 4 + 3} = 4$

Let given point be $P (5, 1)$
Now,
$C P = \sqrt{64 + 1} = \sqrt{65}$

Hence, length of required tangent is
$= \sqrt{C P^{2} - r^{2}} = \sqrt{65 - 16} = \sqrt{49} = 7$

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