The equation of the tangent to the circle x^2 | vedclass

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(D) The circle $x^2+y^2=(8)^2$ has radius $r=8$ and center $(0,0)$.The point $P$ on the circle corresponding to the parameter $	heta = \frac{2\pi}{3}

Vedclass · Accepted Answer

$x-\sqrt{3}y+16=0$

Vedclass · Answer

(D) The circle $x^2+y^2=(8)^2$ has radius $r=8$ and center $(0,0)$.The point $P$ on the circle corresponding to the parameter $	heta = \frac{2\pi}{3}$ has coordinates:$P \equiv (8 \cos \frac{2\pi}{3}, 8 \sin \frac{2\pi}{3}) = (8 	imes -\frac{1}{2}, 8 	imes \frac{\sqrt{3}}{2}) = (-4, 4\sqrt{3})$.The equation of the tangent to the circle $x^2+y^2=r^2$ at point $(x_1, y_1)$ is $xx_1 + yy_1 = r^2$.Substituting $(x_1, y_1) = (-4, 4\sqrt{3})$ and $r^2 = 64$:$-4x + 4\sqrt{3}y = 64$.Dividing by $-4$:$x - \sqrt{3}y = -16 \Rightarrow x - \sqrt{3}y + 16 = 0$.

The equation of the tangent to the circle $x^2+y^2=64$ at the point $P\left(\frac{2\pi}{3}\right)$ is

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