Question

In the given figure, line l touches the circle with centre O at point P. Q is the mid point of radius OP. RS is a chord through Q such that chords RS || line l. If RS = 12 find the radius of the circle.

Answer 1

It is given that line l touches the circle with centre O at point P.

∴ ∠OPX = 90º (Tangent at any point of a circle is perpendicular to the radius through the point of contact)

Join OR.

Now, chord RS || l.

∴ ∠OQR = 90º (Pair of corresponding angles)

⇒ Q is the mid-point of RS. (Perpendicular drawn from the centre of a circle on its chord bisects the chord)

So, RQ = QS = $\frac{12}{2}$ = 6 units

Let r be the radius of the circle.

∴ OR = PQ = r (Radii of the circle)

Q is the mid-point of OP.

∴ OQ = PQ = $\frac{r}{2}$

In right ∆OQR,

$$\begin{gathered} {\mathrm{OR}}^2={\mathrm{OQ}}^2+{\mathrm{RQ}}^2 \\ \Rightarrow r^2={\left(\frac{r}{2}\right)}^2+{\left(6\right)}^2 \\ \Rightarrow r^2-\frac{r^2}{4}=36 \\ \Rightarrow \frac{3r^2}{4}=36 \\ \Rightarrow r=\sqrt{\frac{36\times 4}{3}}=\sqrt{48}=4\sqrt{3}\mathrm{units} \end{gathered}$$
Thus, the radius of the circle is $4\sqrt{3}$ units.