Let A be one point of intersection of two intersecting circles with centres O and Q. The tangents at A to the two circles meet the circles again at B and C,  respectively. Let the point P be located so that AOPQ is a parallelogram. Prove that P is the circumcentre of the triangle ABC in figure.
(Hint: AQ ⊥ AB and AQ || OP. Then OP ⊥ AB and is also bisector of AB. Similarly,  PQ is perpendicular bisector of AC.)

Given: Two circles with centre O and Q intersect at A. The tangents at A to the two circles meet the circles again at B and C, respectively. AOQP is a parallelogram.
To prove: P is the circumcentre of the triangle ABC.
Proof:
AQ ⊥ AB and AQ || OP
Therefore OP ⊥ AB
Also OP bisects AB as the line drawn from the centre to the chord bisects the chord.
Hence OP is the perpendicular bisector of AB.
Similarly PQ is the perpendicular bisector of AC.
Since the perpendicular bisectors intersect at P. P is the circumcentre of the triangle ABC.