(D) We know that the radius of a circle is perpendicular to the tangent at the point of contact.Therefore,$OP \perp PQ$.This forms a right-angled tria

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(D) We know that the radius of a circle is perpendicular to the tangent at the point of contact.Therefore,$OP \perp PQ$.This forms a right-angled triangle $\triangle OPQ$,where $OP = 5 \, cm$ (radius) and $OQ = 12 \, cm$ (hypotenuse).By applying the Pythagoras theorem in $\triangle OPQ$:$OP^2 + PQ^2 = OQ^2$$5^2 + PQ^2 = 12^2$$25 + PQ^2 = 144$$PQ^2 = 144 - 25$$PQ^2 = 119$$PQ = \sqrt{119} \, cm$.

Class 10 Mathematics Circles Textbook - Circles

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$A$ tangent $PQ$ at a point $P$ of a circle of radius $5 \, cm$ meets a line through the centre $O$ at a point $Q$ so that $OQ = 12 \, cm$. The length of $PQ$ is (in $cm$):

A
$12$
B
$13$
C
$8.5$
D
$\sqrt{119}$

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$A$ tangent $PQ$ at a point $P$ of a circle of radius $5 \, cm$ meets a line through the centre $O$ at a point $Q$ so that $OQ = 12 \, cm$. The length of $PQ$ is (in $cm$):

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Fill in the blanks:
$(i)$ $A$ tangent to a circle intersects it in $.........$ point$(s)$.
$(ii)$ $A$ line intersecting a circle in two points is called a $.........$