# Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre

**Solution:**

- Let us consider O as the centre point of the circle.
- Let P be a point outside the circle from which two tangents PA and PB are drawn to the circle. Which touches the circle at point A and B respectively
- Draw a line segment between points A and B such that it subtends ∠AOB at centre O of the circle.

According to Theorem 10.1: The tangent at any point of a circle is always perpendicular to the radius through the point of contact.

∴ ∠OAP = ∠OBP = 90° --- Equation (i)

In a quadrilateral, the sum of 4 angles is 360°

∴ In OAPB,

∠OAP + ∠APB + ∠PBO + ∠BOA = 360°

Using Equation (i), we can write the above equation as

90° + ∠APB + 90° + ∠BOA = 360°

∠APB + ∠BOA = 360° - 180°

∴ ∠APB + ∠BOA = 180°

Where,

∠APB = Angle between the two tangents PA and PB from external point P.

∠BOA = Angle subtended by the line segment joining the point of contact at the centre.

Hence, proved the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre.

**Video Solution:**

## Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre

### NCERT Solutions Class 10 Maths - Chapter 10 Exercise 10.2 Question 10:

Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre

It has been proved that the angle between the two tangents drawn from an external point to a circle, that is, ∠APB is supplementary to the angle subtended by the line segment joining the point of contact at the centre, that is, ∠AOB