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

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Solution

The correct option is C 40
The equation of circle is $x^{2} + y^{2} - 4 x - 6 y + 3 = 0$ .

The length of the tangent from point $(3, - 4)$ is,
$\sqrt{x_{1}^{2} + y_{1}^{2} + g x_{1} + h y_{1} - r}$

$\sqrt{{(3)}^{2} + {(- 4)}^{2} - 4 (3) - 6 (- 4) + 3}$

$= \sqrt{40}$

The square of this length of the tangent is $40$ .

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The square of the length of the tangent from (3,−4) to the circle x2+y2−4x−6y+3=0 is

The correct option is C 40The equation of circle is x2+y2−4x−6y+3=0. The length of the tangent from point (3,−4) is,√x21+y21+gx1+hy1−r √(3)2+(−4)2−4(3)−6(−4)+3 =√40 The square of this length of the tangent is 40.

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