# Difference between revisions of "Circles Tangents Problems"

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'''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)'''<br> | '''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)'''<br> | ||

[[File:fig3.png|200px]] | [[File:fig3.png|200px]] | ||

+ | ==Interpretation of problem== |

## Revision as of 06:54, 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

- The radii of a circle are equal
- Properties of isosceles triangle.
- SAS postulate
- Properties of congruent triangles.

## Prerequisite knowledge

- The radii of a circle are equal.
- In an isosceles triangle angles opposite to equal sides are equal.
- 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)**