Circles Tangents Problems

=Problem 1= Tangents AP and AQ are drawn to circle with centre O, from an external point A. Prove that ∠PAQ=2.∠ OPQ

Interpretation of the problem
O is the centre of the circle and tangents AP and AQ are drawn from an external point A. OP and OQ are the radii. The students have to prove thne angle PAQ=twise the angle OPQ.

Concepts used

 * 1) The radii of a circle are equal.
 * 2) In any circle the radius drawn at the point of contact is perpendicular to the tangent.
 * 3) The tangent drawn from an external point to a circle a] are equal b] subtend equal angle at the centre c] are equally inclined to the line joining the centre and extrnal point.
 * 4) Properties of isoscles triangle.
 * 5) Properties of quadrillateral ( sum of all angles) is 360 degrees
 * 6) Sum of three angles of triangle is 180 degrees.

Algorithm
OP=OQ  radii of the same circle OA is joined. In quadrillateral APOQ , ∠APO=∠AQO=$$90^{0}$$ [radius drawn at the point of contact is perpendicular to the tangent] ∠PAQ+∠POQ=$$180^{0}$$ Or, ∠PAQ+∠POQ=$$180^{0}$$ ∠PAQ = $$180^{0}$$-∠POQ --1 Triangle POQ is isoscles. Therefore ∠OPQ=∠OQP ∠POQ+∠OPQ+∠OQP=$$180^{0}$$ Or ∠POQ+2∠OPQ=$$180^{0}$$ 2∠OPQ=$$180^{0}$$- ∠POQ  --2 From 1 and 2 ∠PAQ=2∠OPQ