The length of the tangent from $(- 3, 1)$ to the circle $3 x^{2} + 3 y^{2} - 5 x - 6 y - 12 = 0$ is

3

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\sqrt{3}

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4

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\sqrt{5}

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Solution

The correct option is A $3$
Coordinates of the point $P (x_{1}, y_{1}) = (- 3, 1)$ and equation of the circle is $3 x^{2} + 3 y^{2} - 5 x - 6 y - 12 = 0$
The given equation may be written as $x^{2} + y^{2} - \frac{5}{3} x - 2 y - 4 = 0$ ...(i)
We know that the standard equation of the circle is $x^{2} + y^{2} + 2 g x + 2 f y + c = 0.$
Comparing Eq. (i) with the standard equation, we get $g = \frac{5}{6}, f = - 1$ and $c = - 4.$
We also know that the length of the tangent from the given point to the circle
$= \sqrt{x_{1}^{2} + y_{1}^{2} + 2 g x_{1} + 2 f y_{1} + c} = \sqrt{9 + 1 + 5 - 2 - 4} = \sqrt{9} = 3.$

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