Question

Ok, solve it at least ​

Answer 1

$\textbf{Answer is in Attachment !}$

Answer 2

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.