Mathematics
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.
Answer
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.
We know that,
The tangent at any point of a circle is always perpendicular to the radius through the point of contact.
∴ ∠OAP = ∠OBP = 90° …………(1)
In a quadrilateral, the sum of interior angles is 360°.
In quadrilateral OAPB,
⇒ ∠OAP + ∠APB + ∠PBO + ∠BOA = 360°
Using Equation (1), 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.
Related Questions
In the given figure, XY and X'Y' are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X'Y' at B. Prove that ∠AOB = 90°.
A quadrilateral ABCD is drawn to circumscribe a circle. Prove that :
AB + CD = AD + BC
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 in figure. Find the sides AB and AC.
Prove that the parallelogram circumscribing a circle is a rhombus.