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 interior 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 AB joining the point of contacts 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.
☛ Check: Class 10 Maths NCERT Solutions Chapter 10
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
Summary:
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. Thus, ∠APB + ∠BOA = 180°.
☛ Related Questions:
- Prove that the parallelogram circumscribing a circle is a rhombus.
- A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see Fig. 10.14). Find the sides AB and AC.
- Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the center of the circle.
- From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is(A) 7 cm(B) 12 cm(C) 15 cm(D) 24.5 cm