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

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

Geogebra file
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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 =Problem-2= In the figure two circles touch each other externally at P. AB is a direct common tangent to these circles. Prove that a). Tangent at P bisects AB at Q b). ∠APB=90° (Exescise-15.2, B.3)