Difference between revisions of "Circles Tangents Problems"
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Circles <math>C_{1}</math> and <math>C_{2}</math> touch internally at a point A and AB is a chord of the circle<math>C_{1}</math> intersecting <math>C_{2}</math> at P, Prove that AP= PB.<br> | Circles <math>C_{1}</math> and <math>C_{2}</math> touch internally at a point A and AB is a chord of the circle<math>C_{1}</math> intersecting <math>C_{2}</math> at P, Prove that AP= PB.<br> | ||
[[Image:problem 3 on circle.png|300px]] | [[Image:problem 3 on circle.png|300px]] | ||
− | Concepts used | + | ==Concepts used== |
1. The radii of a circle are equal | 1. The radii of a circle are equal | ||
2.Properties of isosceles triangle. | 2.Properties of isosceles triangle. | ||
3.SAS postulate | 3.SAS postulate | ||
4.Properties of congruent triangles. | 4.Properties of congruent triangles. | ||
− | Prerequisite knowledge | + | ==Prerequisite knowledge== |
1. The radii of a circle are equal. | 1. The radii of a circle are equal. | ||
2. In an isosceles triangle angles opposite to equal sides are equal. | 2. In an isosceles triangle angles opposite to equal sides are equal. | ||
3.All the elements of congruent triangles 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 ] |
Revision as of 11:45, 13 August 2014
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.
Geogebra file
Concepts used
- The radii of a circle are equal.
- In any circle the radius drawn at the point of contact is perpendicular to the tangent.
- 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.
- Properties of isoscles triangle.
- Properties of quadrillateral ( sum of all angles) is 360 degrees
- 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= [radius drawn at the point of contact is perpendicular to the tangent]
∠PAQ+∠POQ=
Or, ∠PAQ+∠POQ=
∠PAQ = -∠POQ ----------1
Triangle POQ is isoscles. Therefore ∠OPQ=∠OQP
∠POQ+∠OPQ+∠OQP=
Or ∠POQ+2∠OPQ=
2∠OPQ=- ∠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
- In the given figure two circles touch externally.
- AB is the direct common tangent to these circles.
- PQ is the transverse common tangent drawn to these circles at point P.
- Using the tangent properties students have to show AQ=BQ and ∠APB=90°
Concepts used
- 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. - Angle subtended by equal sides are equal.
- Axiom-1:- "Things which are equal to same thing are equal"
[Click here for geogebra animation]
Algorithm
In the above figure AB is direct common tangent to two circles and PQ is the Transverse common tangent.
'Step-1'Bold 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 and touch internally at a point A and AB is a chord of the circle intersecting 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 ]