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(A) The radius of the circle is $OP = 9$. Since $PQ$ is a tangent to the circle at point $P$,the radius $OP$ is perpendicular to the tangent $PQ$ at t

Vedclass · Accepted Answer

$41$

Vedclass · Answer

(A) The radius of the circle is $OP = 9$. Since $PQ$ is a tangent to the circle at point $P$,the radius $OP$ is perpendicular to the tangent $PQ$ at the point of contact. Therefore,$\angle OPQ = 90^{\circ}$. In the right-angled triangle $	riangle OPQ$,by the Pythagorean theorem,we have: $OQ^2 = OP^2 + PQ^2$. Substituting the given values: $OQ^2 = 9^2 + 40^2$. $OQ^2 = 81 + 1600$. $OQ^2 = 1681$. Taking the square root of both sides: $OQ = \sqrt{1681} = 41$. Thus,the length of $OQ$ is $41$.

$A$ line passing through the centre of the circle $O$ intersects the tangent to the circle at $Q$. $P$ is the point of contact. If the radius of the circle is $9$ and $PQ = 40$,then find $OQ$.

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