Question

In the given figure, seg MN is a chord of a circle with centre O. MN = 25, L is a point on chord MN such that ML = 9 and d(O,L) = 5. Find the radius of the circle.

Answer 1

seg MN is a chord of a circle with centre O.

Draw OP ⊥ MN and join OM.



MP = PN = $\frac{\mathrm{MN}}{2}=\frac{25}{2}$ units (Perpendicular drawn from the centre of a circle on its chord bisects the chord)

∴ LP = MP − ML = $\frac{25}{2}-9=\frac{7}{2}$ units

In right ∆OPL,

$$\begin{gathered} {\mathrm{OL}}^2={\mathrm{LP}}^2+{\mathrm{OP}}^2 \\ \Rightarrow \mathrm{OP}=\sqrt{{\mathrm{OL}}^2-{\mathrm{LP}}^2} \\ \Rightarrow \mathrm{OP}=\sqrt{5^2-{\left(\frac{7}{2}\right)}^2} \\ \Rightarrow \mathrm{OP}=\sqrt{25-\frac{49}{4}} \\ \Rightarrow \mathrm{OP}=\sqrt{\frac{51}{4}}=\frac{1}{2}\sqrt{51}\mathrm{units} \end{gathered}$$
In right ∆OPM,

$$\begin{gathered} {\mathrm{OM}}^2={\mathrm{MP}}^2+{\mathrm{OP}}^2 \\ \Rightarrow \mathrm{OM}=\sqrt{{\left(\frac{25}{2}\right)}^2+{\left(\frac{\sqrt{51}}{2}\right)}^2} \\ \Rightarrow \mathrm{OM}=\sqrt{\frac{625+51}{4}} \\ \Rightarrow \mathrm{OM}=\sqrt{\frac{676}{4}} \\ \Rightarrow \mathrm{OM}=\sqrt{169}=13\mathrm{units} \end{gathered}$$
Thus, the radius of the circle is 13 units.