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प्रश्न
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.
उत्तर
seg MN is a chord of a circle with centre O.
Draw OP ⊥ MN and join OM.
MP = PN = `(MN)/2 = 25/2`units (Perpendicular drawn from the centre of a circle on its chord bisects the chord)
∴ LP = MP − ML = `25/2-9=7/2`units
In right ∆OPL,
\[{OL}^2 = {LP}^2 + {OP}^2 \]
\[ \Rightarrow OP = \sqrt{{OL}^2 - {LP}^2}\]
\[ \Rightarrow OP = \sqrt{5^2 - \left( \frac{7}{2} \right)^2}\]
\[ \Rightarrow OP = \sqrt{25 - \frac{49}{4}}\]
\[ \Rightarrow OP = \sqrt{\frac{51}{4}} = \frac{1}{2}\sqrt{51} \] units
In right ∆OPM,
\[{OM}^2 = {MP}^2 + {OP}^2 \]
\[ \Rightarrow OM = \sqrt{\left( \frac{25}{2} \right)^2 + \left( \frac{\sqrt{51}}{2} \right)^2}\]
\[ \Rightarrow OM = \sqrt{\frac{625 + 51}{4}}\]
\[ \Rightarrow OM = \sqrt{\frac{676}{4}}\]
\[ \Rightarrow OM = \sqrt{169} = 13 \] units
Thus, the radius of the circle is 13 units.
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