Question
In a circle with centre $P ,$ chord $AB$ is parallel to a tangent and intersects the radius drawn from the point of contact to its midpoint. If $AB = 16\sqrt3$ then find the radius of the circle
✓
Answer
Given: Chord $A B||$ tangent $X Y$
$A B=16 \sqrt{3}$ units
$PQ$ is radius of the circle.
$P C=C Q$
To find: Radius of the circle, $i.e.,I(PQ)$
Construction: Draw seg $PB.$
In given figure, $ \angle P Q Y=90^{\circ}..(i) [$Tangent theorem$]$
Chord $A B \|$ line $X Y$
$\therefore \angle PCB \cong \angle PQY$
$\therefore \angle PCB =90^{\circ}(ii) ...[$From $(i)]$
Now $C B=\frac{1}{2} A B$
$ \begin{array}{l} \therefore C B=\frac{1}{2} \times 16 \sqrt{3} \ldots . .\left[\begin{array}{c} \text { A perpendicular drawn from the } \\ \text { centre of a circle on its chord } \\ \text { bisects the chord } \end{array}\right] \\\end{array}$
$C B=8 \sqrt{3}....(iii) $
Let the radius of the circle be $x$ units
$\therefore PQ = x$
$\therefore PC =\frac{1}{2} PQ \ldots \ldots . .[ PC = CQ , P - C - Q ]$
$\therefore PC =\frac{1}{2} x \ldots \ldots . (v)$
$\text { In } \triangle PCB ,$
$\angle PCB =90^{\circ} \ldots . .[$From $(ii)] $
$\therefore PB P ^2= PC ^2+ CB ^2 \ldots . .[$ Pythagoras theorem $]$
$\therefore x ^2=\left(\frac{1}{2} x\right)^2+(8 \sqrt{3})^2 \ldots . . .[$ From $(iii), (iv)$ and $( v )]$
$\therefore x ^2=\frac{x^2}{4}+64 \times 3$
$\therefore 4 x^2=x^2+192$
$\therefore 4 x ^2- x ^2=192$
$\therefore 3 x^2=192$
$\therefore x ^2=\frac{192}{3}$
$\therefore x^2=64$
$\therefore x =8$ units $[$Taking square root of both sides$]$
$\therefore$ The radius of the circle is $8$ units.
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