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. a) AQ=QP and BQ=QP  (Tangents drawn from external point are equal) By axiom-1, AQ=BQ ∴tangent at P bisects AB at Q. b) Let ∠QBP=x˚

∴∠QPB=x˚ (∵PQ=BQ) Now Let ∠PAQ=y˚ ∠QPA=y˚ (∵ PQ=AQ) ∴In △PAB

∠PAB+∠PBA+∠APB=180˚

y˚+x˚+(x˚+y˚)=180˚ 2x˚+2y˚=180˚ 2(x˚+y˚)=180˚ x˚+y˚=90˚ ∴ ∠APB=90˚

=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)

Interpretation of problem

 * 1) In the quadrilateral ABCD sides BC, DC & QB are given.
 * 2) AD⊥DC.
 * 3) Asked to find the radius OS or OP

Concepts used

 * 1) Tangents drawn from an external point to a circle are equal.
 * 2) In a quadrilateral, if all angles are equal  and  a pair of adjacent sides are equal then it is a square
 * 3) In a circle, the radius drawn at the point of contact is perpendicular to the tangent

Algorithm
In the fig BC=38 cm  and  BQ=27 cm BQ=BR=27 cm  (∵ by concept 1) ∴CR=BC-BR=38-27=11 cm  CR=SC=11 cm                                                             (∵ by concept 1) DC=25 cm ∴ DS=DC-SC=25-11=14 cm  DS=DP=14 cm                                                               (∵ by concept 1) Also AD⊥DC, OP ⊥ AD and OS ⊥ DC ∠D=∠S=∠P=90˚ ⇒ ∠O=90˚ ∴ DSOP is a Square SO=OP=14 cm      hence Radius of given circle is 14 cm

=Problem 5 [Ex-15.2-B8]= A circle is touching the side BC of △ABC at P. AB and AC when produced are touching the circle at Q and R respectively. Prove that AQ =$$\frac{1}{2}$$ [perimeter of △ABC].

Algorithm
In the figure AQ, AR and BC are tangents to the circle with center O. BP=BQ and PC=CR (Tangents drawn from external point are equal) --                       (1)

Perimeter of △ABC=AB+BC+CA =AB+(BP+PC)+CA =AB+BQ+CR+CA 	             --             (From eq-1) =(AB+BQ)+(CR+CA) =AQ+AR		         	-            (From fig) =AQ+AQ		        	-- ---              (∵AQ=AR) =2AQ

∴ AQ = $$\frac{1}{2}$$ [perimeter of △ABC]

=Problem-6 [Ex-15.4-B3]= In circle with center O, diameter AB and a chord AD are drawn. Another circle drawn with OA as diameter to cut AD at C. Prove that BD=2OC.

Algorithm
In figure, AB is the diameter of circle $$C_{1}$$ and AO is the diameter of the circle $$C_{2}$$ in △ADB and △ACO ∠ADB=90° and ∠ACO=90° [∵angles in the semi circles] ∠DAB=∠CAO  [∵common angles] ∴△ADB∼△ACO  [∵equiangular triangles are similar] ∴$$\frac{AB}{OA}$$=$$\frac{BD}{OC}$$=$$\frac{AD}{AC}$$   [∵corresonding sides of a similar triangles are proportional] But AB=2OA1 (∵diameter is twice the radius of a cicle) $$\frac{AB}{OA}$$=$$\frac{BD}{OC}$$ from (1) $$\frac{2OA}{OA}$$=$$\frac{BD}{OC}$$ ∴BD=2OC