The length of the tangent from the point (4, | vedclass

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(A) 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 +

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$\sqrt{13}$

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(A) 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 circle equation $x^2 + y^2 + 2x - 6y - 6 = 0$,we have $g = 1$,$f = -3$,and $c = -6$.The point is $(4, 5)$,so $x_1 = 4$ and $y_1 = 5$.Length of the tangent $= \sqrt{4^2 + 5^2 + 2(4) - 6(5) - 6}$$= \sqrt{16 + 25 + 8 - 30 - 6}$$= \sqrt{49 - 36}$$= \sqrt{13}$.

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

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