6.14 Touching Circles

 

Theorem: If two Circles touch each other internally or externally, the point of contact and the centers of the circles are collinear.

 

Data: Two circles with centers A and B each other externally at point P (Figure 1)) or internally (Figure 2).

 

 

 

 

 

 

 

 

 

 

 

 

To prove: A, B and P are collinear

Construction: Draw the common tangent RPQ at P. Join AP and BP

 

Proof: (When circles touch externally)

Step

Statement

Reason

1

APQ = 900 =BPQ

RQ is tangent to the circles at P, AP and BP are radii

 

2

APQ+BPQ = 1800

From step 1

 

3

APB is a straight line

Angles APQ andBPQ is a linear pair

A, B and P are collinear

 

Proof: (When circles touch internally)

Step

Statement

Reason

1

AP and BP are perpendicular to same line RQ

RQ is tangent to circles at P, AP and BP are radii

 

2

B is a point on line AP

 

3

ABP is a straight line

Step 2

 

A, B and P are collinear

 

6.14 Problem 1: A straight line drawn through the point of contact of two circles whose centers are A and B, intersect the circles at P and Q

respectively. Show that AP and BQ are parallel.

 

In the adjoining figure circles with origins A and B, touch externally at M. We need to prove that AP || BQ.

Step

Statement

Reason

1

AM=AP

Radii of same circle

2

APM = AMP

2 sides are equal

3

AMP= QMB

Vertically opposite angles

4

BM=BQ

Radii of same circle

5

QMB = BQM

2 sides are equal

6

APM = BQM

Steps 2, 3 and 5

7

AP || BQ

APM, BQM are alternate angles

 

6.14 Problem 2: In the given figure AB is line segment, M is the midpoint of AB. 2 semi circles with AM and MB as diameters are drawn on the

line AB. A circle with center as O touches all the three circles.

Prove that the radius of this circle is (1/6)AB

 

Let the radius of this circle be x : OR=OP =x, and AB=a

CP = CM= a/4 and MR=a/2

Step

Statement

Reason

1

OMC is right angled triangle

OM is tangent at M to circle C1

 

2

OC2 = MC2+OM2

Pythagoras theorem on OMC

 

3

LHS = (x+(a/2))2 = x2+ax/2+ (a2/16)

OC = OP+PC = x+(a/4)

 

4

RHS = (a2/16)+ (a2/4)-ax+ x2

MC=a/4,OM = MR-OR=a/2-x

 

5

x2+ax/2+ (a2/16) =(a2/16)+ (a2/4)-ax+ x2

Equating LHS and RHS

6

3ax/2=(a2/4)

On transformation

7

x = a/6

 


 

6.14 Theorem: The tangents drawn to a circle from an external point are

(i)                 Equal

(ii)               Equally inclined to the line joining the external point and the center

(iii)              Subtend equal angles at the center

 

Data: PA and PB are tangents from P to the circle with origin at O

To Prove

(i)                 PA=PB

(ii)               APO= BPO

(iii)              AOP= BOP

 

Proof:

 

Step

Statement

Reason

1

OA = OB

Radii of same circle

2

OAP= OBP= 900

PA and PB are tangents at A and B and AO and BO are radii

3

OP is common side of AOP, BOP

 

4

AOP BOP

SAS Postulate of Right angled triangle

5

PA=PB

Properties of congruent triangles

6

APO= BPO

Properties of congruent triangles

7

AOP= BOP

Properties of congruent triangles

 

6.14 Problem 3: In the figure, XY and PC are common tangents to 2 touching circles. Prove that XPY = 90

Step

Statement

Reason

1

CX= CP

CX and CP are tangents from C

2

CXP =CPX =x0

2 sides are equal

3

CY =CP

CY and CP are tangents from C

4

PYC =CPY =y0

2 sides are equal

5

CXP + XPC + CPY +PYC = 1800

Sum of all angles in a triangle

6

i.e. x0+x0+y0+y0= 1800

 

7

2(x0+y0)= 1800

 

8

i.e. (x0+y0) =XPY = 900

 


6.14 Problem 4: Tangents PQ and PR are drawn to the circle from an external point P. If PQR = 600 prove that the length of the

chord QR = length of the tangent

 

Step

Statement

Reason

1

PQ=PR

PQ and PR are tangents from P

 

2

PQR =PRQ

2 sides are equal

 

3

PQR =600

Given

 

4

PQR =PRQ = 600

Step 2

 

5

PQR is an equilateral triangle

All angles are = 600

6

PQ=PR=QR

 

 

6.14 Problem 5: In the figure PQ and PR are tangents to the circle with Center O. If QPR= 900. Show that PQOR is a square.

Step

Statement

Reason

1

OQP= 900 =ORP

PQ and PR are tangents from P

 

2

QPR=900

Given

3

OQ is parallel to PR

Corresponding angles are 900

 

4

QOR =3600-OQP-QPR -ORP =

3600-900-900-900

Property of quadrilateral

5

OR is parallel to QP

Corresponding angles are 900

 

6

PQOR is a parallelogram

 

7

PQOR is a square

OQ=OR(radii)

 

 

6.14 Problem 6: In the figure, AT and BT are tangents to a circle with center O. Another tangent PQ is drawn such that TP=TQ.

Show that TAB ||| TPQ

 

 

Step

Statement

Reason

1

AT=BT

TA and TB are tangents from an external point T

2

TAB=TBA

2 sides are equal

3

PT=QT

TP and TQ are tangents from an external point T

4

TPQ=TQP

2 sides are equal

5

ATB= 1800- (TAB+TBA)=

1800- 2TAB

InTAB, sum of all the angles = 1800

6

ATB= 1800- (TPQ+TQP)=

1800- 2TPQ

InTPQ, sum of all the angles = 1800

7

TAB =TPQ

Equate RHS of steps 5, 6

8

TAB =TPQ=TQP =TBA

Steps 7, 4, 2

9

TAB ||| TPQ

Triangles are equiangular

 

6.14 Problem 7: In the given figure, tangents are drawn to the circle from external points A, B and C. Prove that

1) AP+BQ+CR = BP+CQ+AR and AP+BQ+CR = 1/2 *perimeter of ABC.

2) If AB=AC, prove that BQ=QC

 

Step

Statement

Reason

1

PA=AR

Tangents to circle from A

2

BQ=BP

Tangents to circle from B

3

CR=CQ

Tangents to circle from C

4

PA+BQ+CR=AR+BP+CQ

Addition of steps 1 to 3

5

AB=AP+PB, BC=BQ+QC, AC=AR+RC

 

6

AB+BC+AC = PA+BQ+CR +AR+BP+CQ

Addition of step 5

7

= 2 (AP+BQ+CR) = Perimeter of ABC

From Step 4

Second part

8

AB=AC

Given

9

AP+PB=AR+RC

 

10

PB=RC

Since AP = AR, Step 1

11

BQ=CQ

Steps 2 and 3

6.14 Problem 8: TP and TQ are tangents drawn to a circle with O as center.

Show that

 

1. OT is perpendicular bisector of PQ

2. PTQ =2OPQ

 

 

Step

Statement

Reason

Consider TPR and TQR

1

TP=TQ, PTR=QTR

6.14 Theorem(TP and TQ are tangents)

2

TR is common

 

3

TPR TQR

SAS Postulate on congruence

4

PR=RQ and PRT=QRT

Corresponding sides are equal

5

PRT = 900

Two equal angles on a straight line

Second Part

6

PTR +RPT = 900

Sum of two angles in a right angled triangle PRT

7

OPT =900=OPR+RPT

PT is tangent and OP is radius P = 900

8

PTR =OPR

Steps 6 and 7

9

PTQ = 2 PTR

Step 1

10

= 2 OPR

Step 8

 

Note: Above problem can also be solved by using the properties: OP=OQ and OPT = 900

 

6.14 Summary of learning

 

 

No

Points to remember

1

The tangents drawn to a circle from an external point are

-equal,

-equally inclined to the line joining the external point and circle,

-subtend equal angles at the center.

 

 

Additional Points:

 

6.14. Theorem 1: A tangent at any point on the circle is perpendicular to the radius through that point.

6.14. Theorem 2: The line perpendicular to a radius at its outer end is a tangent to the circle.

Above two theorems can be proved by logical reasoning (First we make an assumption that the theorem is not true. Subsequently,

because of a contradiction, we conclude that our assumption is wrong. Then by logical reasoning we conclude that theorem is true.)

 

6.14.Theorem 3: If a chord(AB) and a tangent(PT) intersect externally, then the product of lengths of the segments of the chord (PA.PB)

is equal to the square of the length of the tangent(PT2)from the point of contact(T) to the point of intersection (P).

Given: PT is tangent, AB is chord.

To prove: PA.PB = PT2

Construction: Join O to the mid point M of AB, Join OA.

 

Step

Statement

Reason

1

PA = PM-AM

Construction

2

PB = PM+MB

Construction

3

= PM+AM

MB=AM(Construction)

4

PA.PB = (PM-AM)*(PM+AM)

Product of Steps 1 and 3

5

= PM2-AM2

Expansion

6

PM2 = PO2-OM2

Pythagoras theorem on POM

7

AM2 = AO2-OM2

Pythagoras theorem on AOM

8

PA.PB = PO2 - AO2

Substitute results from Step 6 and 7 in Step 5

9

= PO2-TO2

AO=TO(Radii)

10

PA.PB = PT2

Pythagoras theorem on PTO

 

 

 

 

 

 

 

 

 

 

 

 

 

 

6.14. Theorem 4 (Tangent-Secant theorem or Alternate Segment theorem): The angle between a tangent (PQ) and a chord (AB)

through the point of contact is equal to an angle in the alternate segment.

Given: PQ is a tangent at A to the circle with O as center. AB is a chord.

To prove: If C is a point on a major arc and D is a point on a minor arc with respect to the chord AB then

BAQ = ACB and PAB = ADB

Construction: Join OB

Step

Statement

Reason

 

 

1

OAQ = 900

The line drawn from point of contact to center is at right angle to the tangent

2

OAB = 900 - BAQ

Split OAQ

3

BAO = ABO

OA=OB(Radii) hence OAB is an isosceles triangle

4

AOB = 1800 - 2OAB

Sum of all angles in AOB is 1800 and step 3

5

= 1800 2 (900 - BAQ)

Step 2

6

= 2BAQ

Simplification

7

AOB = 2ACB

Angles on the same chord AB at center and circle

(Refer 6.8.2 Inscribed angle theorem)

8

BAQ = ACB

Steps 6 and 7

9

PAB + BAQ = 1800

Angles on the straight line

10

ADB + ACB = 1800

Opposite angles in a cyclic quadrilateral are supplementary (Refer 6.9.3)

11

PAB = ADB

Steps 8,9 and 10