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)

Interpretation of the problem

 * 1) In the given figure two circles touch externally.
 * 2) AB is the direct common tangent to these circles.
 * 3) PQ is the transverse common tangent drawn to these circles at point P.
 * 4) Using the tangent properties students have to show AQ=BQ and ∠APB=90°

Concepts used
[Click here for geogebra animation]
 * 1) The tangent drawn from an external point to a circle a) are equal b] subtend equal angle at the center  c] are equally inclined to the line joining the center and external point.
 * 2) Angle subtended by equal sides are equal.
 * 3) Axiom-1:- "Things which are equal to same thing are equal"

Algorithm
In the above figure AB is direct common tangent to two circles and PQ is the Transverse common tangent. Step-1Bold text AQ=QP and BQ=QP  (Tangents drawn from external point are equal) By axiom-1, AQ=BQ ∴tangent at P bisects AB at Q. =problem 3 [Ex-15.2 B.7]= Circles $$C_{1}$$  and   $$C_{2}$$   touch internally at a point  A and AB is a chord of the circle$$C_{1}$$     intersecting  $$C_{2}$$  at P, Prove that  AP= PB.

Concepts used

 * 1) The radii of a circle are equal
 * 2) Properties of isosceles triangle.
 * 3) SAS postulate
 * 4) Properties of congruent  triangles.

Prerequisite knowledge

 * 1) The radii of a circle are equal.
 * 2) In an isosceles triangle  angles opposite to equal sides  are equal.
 * 3) All the elements of congruent triangles are  equal.

Algoritham
In ∆AOB AO=BO [Radii of a same circle] ∴ ∠OAB = ∠OBA --I [∆AOB is an isosceles ∆} Then, In ∆AOP and ∆BOP, AO = BO [Radii of a same circle] OP=OP [common side] ∠OAP = ∠OBP    [ from I] ∴ ∆AOP ≅ ∆BOP [SAS postulate] ∴ AP = BP [corresponding sides of congruent triangles ]

=problem-4= In the given Quadrilateral ABCD, BC=38cm , QB=27cm , DC=25cm and AD⊥DC find the radius of the circle.(Ex:15.2. A-6)