Question

In the figure, the diameter $CD$ of a circle with centre $O$ is perpendicular to the chord $AB$. If $AB=12cm$ and $CE=3cm$, find the radius of the circle (in $cm$)

Answer 1

Let us Join $OA$ to make a right-angled triangle.

Since, a perpendicular drawn from the centre of circle to a chord, bisects the chord.

So, $AE=EB=6cm$

Let $r$ be the radius of the circle.
Then, $OA=OC=rcm$

$OE=(OC–CE)=(r–3)cm$.

In right-angled $DOEA$, we have

${OA}^2$ = ${AE}^2$ + ${OE}^2$ [By Pythagoras Theorem]

$r^2$ = $6^2$ + ${(r-3)}^2$ [$OA=rcm,AE=6cm$ and $OE=(r–3)cm$]

$r^2$ = $36+r^2$ $-6r+9$ [Since, $(a-b{)}^2=a^2+b^2-2ab$]

$\Rightarrow 6r=45$ [ $r^2$ gets cancelled]

$\therefore r=7.5$

Hence, the radius of the circle is $7.5cm$.