overline{PA} is a tangent to odot(O, r) drawn | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Since the tangent at any point of a circle is perpendicular to the radius through the point of contact,$\angle OAP = 90^{\circ}$.In right-angled t

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(C) Since the tangent at any point of a circle is perpendicular to the radius through the point of contact,$\angle OAP = 90^{\circ}$.In right-angled triangle $	riangle OAP$,by Pythagoras theorem:$OP^{2} = OA^{2} + AP^{2}$Given $OP = 10$ and $AP = 8$,we have:$10^{2} = OA^{2} + 8^{2}$$100 = OA^{2} + 64$$OA^{2} = 100 - 64 = 36$$OA = \sqrt{36} = 6$Here,$OA$ is the radius $(r)$ of the circle.Diameter of the circle $= 2 	imes 	ext{radius} = 2 	imes 6 = 12$.

$\overline{PA}$ is a tangent to $\odot(O, r)$ drawn from a point $P$ outside a circle. If $OP = 10$ and $AP = 8$,then the diameter of the circle is equal to $\ldots \ldots \ldots \ldots .$

Similar Questions

State 'True' or 'False' and give reasons for your answer.
If the angle between two tangents drawn from a point $P$ to a circle of radius $a$ and center $O$ is $60^{\circ}$,then $OP = a\sqrt{3}$.

$A$ tangent $\overline{PM}$ is drawn from a point $P$ outside $\odot(O, 8)$. $\overline{OP}$ intersects the circle at $N$. If $NP = 2$,then $PM = \ldots$

$A$ circle touches all the sides of a quadrilateral $ABCD$. Then, the quadrilateral $ABCD$ is a:

If a rhombus $ABCD$ is inscribed in a circle with center $O$ and radius $r$,then the rhombus $ABCD$ is $\ldots \ldots$

$\overleftrightarrow{PA}$ and $\overleftrightarrow{PB}$ are tangents to the circle $\odot(O, r)$ at $A$ and $B$ respectively. If $m\angle AOB = 100^\circ$,then $m\angle OPB = \dots$ (in $^\circ$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module