The length of the tangent from (6,8) to the c | vedclass

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Question

(C) Let the point be $P = (6, 8)$ and the equation of the circle be $S: x^2 + y^2 - 4 = 0$. The length of the tangent from a point $(x_1, y_1)$ to a c

Vedclass · Accepted Answer

$4 \sqrt{6}$

Vedclass · Answer

(C) Let the point be $P = (6, 8)$ and the equation of the circle be $S: x^2 + y^2 - 4 = 0$. The length of the tangent from a point $(x_1, y_1)$ to a circle $S = 0$ is given by $\sqrt{S_1} = \sqrt{x_1^2 + y_1^2 - 4}$. Substituting the coordinates $(6, 8)$ into the expression: $	ext{Length} = \sqrt{6^2 + 8^2 - 4} = \sqrt{36 + 64 - 4} = \sqrt{100 - 4} = \sqrt{96}$. Simplifying $\sqrt{96} = \sqrt{16 	imes 6} = 4 \sqrt{6}$. Hence,option $C$ is correct.

The length of the tangent from $(6,8)$ to the circle $x^2+y^2=4$ is

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