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. 

Explanation:- 

Consider a circle with center O and a chord MN. Let OP be perpendicular to MN and OM be another line joining O to M. We need to find the radius of the circle.

By the perpendicular bisector theorem, OP is the perpendicular bisector of MN. Therefore, MP = PN = (MN)/2 = 25/2 units.

We also have ML = 9 units. Thus, LP = MP – ML = 25/2 – 9 = 7/2 units.

Using Pythagoras theorem in right triangle OLP, we get:

${OL}^2 = {LP}^2 + {OP}^2$

Solving for OP, we get:

$OP = \sqrt{{OL}^2 – {LP}^2}$

Substituting the values, we get:

$OP = \sqrt{5^2 – \left( \frac{7}{2} \right)^2} = \sqrt{25 – \frac{49}{4}} = \sqrt{\frac{51}{4}} = \frac{1}{2}\sqrt{51}$ units.

Next, using Pythagoras theorem in right triangle OPM, we get:

${OM}^2 = {MP}^2 + {OP}^2$

Substituting the values, we get:

$OM = \sqrt{\left( \frac{25}{2} \right)^2 + \left( \frac{\sqrt{51}}{2} \right)^2} = \sqrt{\frac{625 + 51}{4}} = \sqrt{\frac{676}{4}} = \sqrt{169} = 13$ units.

Therefore, the radius of the circle is 13 units.