overline{PA} is a tangent to odot(O, 8) drawn | vedclass

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

Question

(D) In $\Delta OAP$,the radius $OA$ is perpendicular to the tangent $PA$ at the point of contact $A$. Therefore,$m\angle OAP = 90^\circ$.Given $m\angl

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(D) In $\Delta OAP$,the radius $OA$ is perpendicular to the tangent $PA$ at the point of contact $A$. Therefore,$m\angle OAP = 90^\circ$.Given $m\angle AOP = 45^\circ$.In $\Delta OAP$,the sum of angles is $180^\circ$.$m\angle OPA = 180^\circ - (90^\circ + 45^\circ) = 180^\circ - 135^\circ = 45^\circ$.Since $m\angle AOP = m\angle OPA = 45^\circ$,$\Delta OAP$ is an isosceles triangle.Therefore,the sides opposite to equal angles are equal: $AP = OA$.Since $OA$ is the radius of the circle,$OA = 8$.Thus,$AP = 8$.

$\overline{PA}$ is a tangent to $\odot(O, 8)$ drawn from a point $P$ outside the circle. If $m\angle AOP = 45^\circ$,then $AP = \ldots$

Similar Questions

In $\Delta ABC$,$\angle B = 90^{\circ}$. The radius of the incircle touching all three sides of the triangle is $\ldots \ldots \ldots \ldots$

Fill in the blanks:
$(i)$ $A$ circle can have $......$ parallel tangents at the most.
$(ii)$ The common point of a tangent to a circle and the circle is called $......$

In a right triangle $ABC$ in which $\angle B = 90^{\circ}$,a circle is drawn with $AB$ as diameter,intersecting the hypotenuse $AC$ at $P$. Prove that the tangent to the circle at $P$ bisects $BC$.

$P$ is a point in the exterior of a circle having centre $O$ and radius $21$. $OP = 25$. $A$ tangent from $P$ touches the circle at $Q$. Find $PQ$.

$P$ is a point in the exterior of $\odot(O, r)$ and the tangents from $P$ to the circle touch the circle at $X$ and $Y$. Find $OP$,if $r = 12$ and $XP = 5$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module