Let `A` be one point of intersection of two intersecting circles with centres `O and Q`. The tangents at `A` to the two circls meet the circles again at `B and C`, respectively. Let the point `P` be located so that `A O P Q` is a parallelogram. Prove that `P` is the circumcentre of the triangle `ABC`.

Explanation

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 center at O and OA is its radius.

∴ `OA _|_AC`

⇒ `PQ _|_ AC` [∵OAQP is a parallelogram ]

⇒PQ is the perpendicular bisector of AC. [∵Q is the centre of the circle]

Similarly, BA is the tangent to the circle at A and AQ is its radius, through A.

∴BA `_|_` AQ

∴BA `_|_` OP [∵AQPO is parallelogram]

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