Theorems which are useful to find solution:
i. The segment joining the centre of a circle and the midpoint of a chord is perpendicular to the chord.
ii. In a right angled triangle, sum of the squares of the perpendicular sides is equal to square of its hypotenuse.
\(QP = 1/2 \) (QR) [P is the midpoint of chord QR]

1/2 × 24 = 12 units
Also, seg OP ⊥ chord QR [The segment joining centre of a circle and midpoint of a chord is perpendicular to the chord]

∴ The radius of the circle is 2√61 units.