Question

In the adjoining 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(0, L) = 5. Find the radius of the circle.

Answer 1

Construction: Draw seg OK ⊥ chord MN. 

Join OM. 

seg OK ⊥chord MN [Construction] 

∴ MK = 1/2 MN [Perpendicular drawn from the centre of the circle to the chord bisects the chord] 

= 1/2 × 25 

= 12.5 units 

MK = ML + LK [M – L – K]

∴ 12.5 = 9 + LK 

∴ LK= 12.5 – 9 = 3.5 units 

In ∆OKL, ∠OKL = 90° 

 ∴ OL = KL + OK [Pythagoras theorem] 

∴ 5² = 3.5² + OK² 

∴ OK² = 25 – 12.25 = 12.75 

Now, in ∆OKM, ∠OKM = 90° 

∴ OM² = OK² + MK²

 = 12.75 + 12.52 

= 12.75 + 156.25 

= 169 

∴ OM = √169

= 13 units [Taking square root of both sides]

 ∴ The radius of the given circle is 13 units.