Let A be one point of …

Given that A be the one of the points 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 and AOPQ is a parallelogram. 

 Now In order to prove that P is the circumcentre of {tex}\Delta{/tex}ABC, it is sufficient to show that P is the point of intersection of perpendicular bisectors of the sides of ∆ ABC  i. e, OP and PQ are perpendicular bisectors of sides AB and AC respectively.
Now, AC is tangent at A to the circle with centre at O and OA is its radius.
{tex}\therefore \quad O A \perp A C{/tex}
{tex}\Rightarrow \quad P Q \perp A C{/tex} {tex}[ \because O A Q P \text { is a parallelogram } \therefore O A \| P Q ]{/tex}
=> PQ is the perpendicular bisector of AC.{tex}[ \because Q \text { is the centre of the circle } ]{/tex}
Similarly, BA is the tangent at A to the circle with centre at Q and AQ is its radius through A.
{tex}\therefore \quad B A \perp A Q{/tex}
{tex}\therefore \quad B A \perp O P{/tex} {tex}[ \because A Q P O \text { is parallelogram } \therefore O P \parallel A Q ]{/tex}
{tex}\Rightarrow{/tex} OP is the perpendicular bisector of AB.
Thus, P is the point of intersection of perpendicular bisectors PQ and PO of sides AC and
AB respectively
Hence, P is the circumcentre of {tex}\Delta A B C{/tex}.