A and B are the centres of two unequal circles which intersect at P and Q respectively as shown in - Maths - Circles

Question

A and B are the centres of two unequal circles
which intersect at P and Q respectively as shown in figure Prove
that âˆ  PAB =
âˆ  QAB Also show that M is the mid point of
PQ and angles at M are right angles

Vimala · Accepted Answer

Since the line segment PQ is drawn connecting the two points where the circles intersect, it is a common chord.

Thus, P and Q lie on both the circles.

Thus, AP = AQ = radius of the bigger circle.

Property of circle:

A line from the centre of a circle to a chord is perpendicular to it and bisects it.

Thus, AM is perpendicular to PQ and AM bisects PQ.

Thus, it has been proven that:

1. M is the mid-point pf PQ. That is, PM=MQ

2. Angles at $∠ P M A = ∠ P M B = 90^{\circ}$ are at right angles.

Thus, we have, AP = AQ (radius of the bigger circles), AM = AM (common side), and PM = MQ.

Therefore, the two triangles PMA and PMB are congruent by SSS congruency.

Hence the $∠ P A B$ and $∠ Q A B$ are also congruent.

Thus, we have, $∠ P A M = ∠ Q A M$

That is, $∠ P A B = ∠ Q A B$

A and B are the centres of two unequal circles which intersect at P and Q respectively as shown in figure. Prove that ∠ PAB = ∠ QAB.Also show that M is the mid point of PQ and angles at M are right angles