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