The angle between the tangents to the circle | vedclass

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Question

(D) The equation of the circle is $x^2+y^2=25$,so the center is $O(0,0)$ and the radius $r=5$.Let $P$ be the point $(1,7)$. The distance $OP = \sqrt{1

Vedclass · Accepted Answer

$\frac{\pi}{2}$

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(D) The equation of the circle is $x^2+y^2=25$,so the center is $O(0,0)$ and the radius $r=5$.Let $P$ be the point $(1,7)$. The distance $OP = \sqrt{1^2+7^2} = \sqrt{1+49} = \sqrt{50} = 5\sqrt{2}$.Let $T$ be the point of tangency. In the right-angled triangle $	riangle OPT$,$\sin(\angle OPT) = \frac{OT}{OP} = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}}$.Therefore,$\angle OPT = 45^{\circ}$.The angle between the two tangents is $2 	imes \angle OPT = 2 	imes 45^{\circ} = 90^{\circ}$ or $\frac{\pi}{2}$.

The angle between the tangents to the circle $x^2+y^2=25$ from the point $(1,7)$ is

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