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(C) The length of the tangent from a point $(x_1, y_1)$ to a circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is given by $\sqrt{x_1^2 + y_1^2 + 2gx_1 + 2fy_1 +

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(C) The length of the tangent from a point $(x_1, y_1)$ to a circle $x^2 + y^2 + 2gx + 2fy + c = 0$ is given by $\sqrt{x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c}$.Given the point $(3, -4)$ and the circle $x^2 + y^2 - 4x - 6y + 3 = 0$.Let $L$ be the length of the tangent.$L^2 = x_1^2 + y_1^2 - 4x_1 - 6y_1 + 3$Substitute $x_1 = 3$ and $y_1 = -4$:$L^2 = (3)^2 + (-4)^2 - 4(3) - 6(-4) + 3$$L^2 = 9 + 16 - 12 + 24 + 3$$L^2 = 25 - 12 + 24 + 3$$L^2 = 13 + 24 + 3 = 40$.Thus,the square of the length of the tangent is $40$.

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

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