P is in the exterior of odot (O, 9). A tangen | vedclass

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(C) Given that the circle has center $O$ and radius $r = 9$. Point $P$ is in the exterior of the circle,and $PT$ is a tangent to the circle at point $

Vedclass · Accepted Answer

$41$

Vedclass · Answer

(C) Given that the circle has center $O$ and radius $r = 9$. Point $P$ is in the exterior of the circle,and $PT$ is a tangent to the circle at point $T$. According to the property of tangents,the radius is perpendicular to the tangent at the point of contact. Therefore,$\angle OTP = 90^\circ$. In the right-angled triangle $	riangle OTP$,by the Pythagorean theorem: $OP^2 = OT^2 + PT^2$ Given $OT = r = 9$ and $PT = 40$. $OP^2 = 9^2 + 40^2$ $OP^2 = 81 + 1600$ $OP^2 = 1681$ $OP = \sqrt{1681} = 41$. Thus,the length of $OP$ is $41$.

$P$ is in the exterior of $\odot (O, 9)$. $A$ tangent from $P$ touches the circle at $T$. If $PT = 40$,then $OP = \ldots$

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