Question

pe questions of 5 marks each.
4. Let A be the point of intersection of two
intersecting circles with centre 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 AABC.​

Answer 1

Answer:

Let A be one point of intersection of two intersecting circle with centres O and Q. The tangent 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.

Step-by-step explanation:

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Answer 2

In order to prove that P is the circumcentre of △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 center at O and OA is its radius.

∴OA⊥AC

⇒PQ⊥AC [∵OAQP is a parallelogram

∴OA∥PQ]

⇒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

∴OP∥AQ]

⇒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 △ABC