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 = 90^{0 }=_{}BPQ 
RQ
is tangent to the circles at P, AP and BP are radii 

2 
_{}APQ+_{}BPQ = 180^{0} 
From
step 1 

3 
APB
is a straight line 
Angles
_{}APQ and_{}BPQ 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 
_{ } 

2 
OC^{2} = MC^{2}+OM^{2} 
Pythagoras theorem on _{}OMC 

3 
LHS = (x+(a/2))^{2} = x^{2}+ax/2+
(a^{2}/16) 
OC
= OP+PC = x+(a/4) 

4 
RHS = (a^{2}/16)+ (a^{2}/4)ax+
x^{2} 
MC=a/4, 

5 
x^{2}+ax/2+ (a^{2}/16)
=(a^{2}/16)+ (a^{2}/4)ax+ x^{2} 
Equating LHS and RHS 

6 
3ax/2=(a^{2}/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)
_{}
(iii)
_{}AOP= _{}BOP
Proof:
Step 
Statement 
Reason 

1 
OA
= 
Radii
of same circle 

2 
_{}OAP= _{}OBP= 90^{0} 
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 
_{} 
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 =x^{0} 
2 sides are equal 

3 
CY =CP 
CY and CP are tangents from C 

4 
_{}PYC =_{}CPY =y^{0} 
2 sides are equal 

5 
_{}CXP + _{}XPC + _{}CPY +_{}PYC = 180^{0} 
Sum of all angles in a triangle 

6 
i.e. x^{0}+x^{0}+y^{0}+y^{0}= 180^{0} 


7 
2(x^{0}+y^{0})= 180^{0} 


8 
i.e. (x^{0}+y^{0}) =_{}XPY = 90^{0} 

6.14 Problem 4: Tangents PQ and PR are drawn to the circle from an
external point P. If _{}PQR = 60^{0} 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 =60^{0} 
Given 

4 
_{}PQR =_{}PRQ = 60^{0} 
Step
2 

5 
PQR
is an equilateral triangle 
All
angles are = 60^{0} 

6 
PQ=PR=QR


6.14 Problem 5: In the figure PQ and PR are tangents to the circle
with Center O. If _{}QPR= 90^{0}. Show that PQOR is a square.
Step 
Statement 
Reason 

1 
_{}OQP= 90^{0} =_{}ORP 
PQ
and PR are tangents from P 

2 
_{}QPR=90^{0} 
Given 

3 
OQ
is parallel to PR 
Corresponding angles are 90^{0} 

4 
_{}QOR =360^{0}_{}OQP_{}QPR _{}ORP = 360^{0}90^{0}90^{0}90^{0} 
Property
of quadrilateral 

5 
OR
is parallel to QP 
Corresponding angles are 90^{0} 

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= 180^{0} (_{}TAB+_{}TBA)= 180^{0}
2_{}TAB 
In_{}TAB, sum of all the angles = 180^{0} 

6 
_{}ATB= 180^{0} (_{}TPQ+_{}TQP)= 180^{0}
2_{}TPQ 
In_{}TPQ, sum of all the angles = 180^{0} 

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 =2_{}OPQ
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 = 90^{0} 
Two
equal angles on a straight line 

Second Part 

6 
_{}PTR +_{}RPT = 90^{0} 
Sum
of two angles in a right angled triangle PRT 

7 
_{}OPT =90^{0}=_{}OPR+_{}RPT 
PT
is tangent and OP is radius _{}P = 90^{0} 

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 = 90^{0}
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(PT^{2})from the point of contact(T) to the point of intersection (P).^{}
Given: PT is
tangent, AB is chord.
To prove: PA.PB = PT^{2}
Construction: Join O to the mid point M of AB, Join
OA.
Step 
Statement 
Reason 

1 
PA = PMAM 
Construction 

2 
PB = PM+MB 
Construction 

3 
= PM+AM 
MB=AM(Construction) 

4 
PA.PB = (PMAM)*(PM+AM) 
Product of Steps 1 and 3 

5 
= PM^{2}AM^{2}^{} 
Expansion 

6 
PM^{2 }= PO^{2}OM^{2} 
Pythagoras theorem on _{}POM 

7 
AM^{2 }= AO^{2}OM^{2} 
Pythagoras theorem on _{}AOM 

8 
PA.PB = PO^{2 } AO^{2} 
Substitute results from Step 6
and 7 in Step 5 

9 
= PO^{2}TO^{2} 
AO=TO(Radii) 

10 
PA.PB = PT^{2} 
Pythagoras theorem on _{}PTO 
6.14. Theorem 4 (TangentSecant 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
Step 
Statement 
Reason 

1 
_{}OAQ = 90^{0} 
The line drawn from point of
contact to center is at right angle to the tangent 

2 
_{}OAB = 90^{0 } _{} BAQ^{} 
Split _{}OAQ 

3 
_{}BAO = _{}ABO 
OA= 

4 
_{}AOB = 180^{0 } 2_{}OAB 
Sum of all angles in _{}AOB is 180^{0 }and step 3 

5 
= 180^{0 }– 2 (90^{0 }_{} BAQ) 
Step 2 

6 
= 2_{}BAQ 
Simplification 

7 
_{}AOB = 2_{}ACB 
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 = 180^{0} 
Angles on the straight line 

10 
_{}ADB + _{}ACB = 180^{0} 
Opposite angles in a cyclic
quadrilateral are supplementary (Refer 6.9.3) 

11 
_{}PAB = _{}ADB 
Steps 8,9 and 10 