CBSE Class 10 Answered

Let us consider a circle centred at point O. Let P be a external point from which two tangents PA and PB are drawn to circle which are touching circle at point A and B respectively and AB is the line segment, joining point of contacts A and B together such that it subtends AOB at centre O of circle.

Now we may observe that:

OA (radius) PA (tangent)

So, OAP = 90^o

Similarly, OB (radius) PB (tangent)

OBP = 90^o

Now in quadrilateral OAPB, sum of all interior angles = 360^o

OAP + APB + PBO +BOA = 360^o

90^o + APB + 90^o + BOA = 360^o

APB +BOA = 180^o

Hence, 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.