SSC math

1. 3. CIRCLE

2. A
C
D
A segment joining two distinct points on a circle is called a chord.
O
CHORD

3. A
C
D
P
Q
O
diameter
A chord passing through the centre of the circle is called a diameter.
DIAMETER

4. A
C
D
P
Q
O
diameter
M
N
secant
A line intersecting a circle in two distinct points is called a secant.
Note :
A secant contains a chord
SECANT

5. O
l
A
A line in the plane of a circle, touching the circle at one point
is called tangent.
Point of contact
B
Tangent segment
A segment joining any point on the tangent and the point of contact
is called a tangent segment.
Remember :
For a tangent segment one
of the endpoints has to be
the point of contact
TANGENT

6. THEOREM
O
A
l
seg OA ^ line l
B
m OAB = 90o
OR
[Radius is perpendicular to the tangent]
Result :
A tangent at any point of a circle is perpendicular to the
radius, through the point of contact.

7. THEOREM
A tangent at any point of a circle is perpendicular to the
radius, through the point of contact.
O
A
l
B
Given : Line l is a tangent to the circle with centre O at
the point of contact A.
To prove : Line l | radius OA.
Proof : Assume that, line l is not perpendicular to
seg OA.
Suppose, seg OB is drawn perpendicular to
line l.
Of course B is not same as A.
Now take a point C on line l such that A-B-
C and BA = BC.
Now in, Δ OBC and Δ OBA
Seg BC = seg BA ….( construction )
OBC = OBA ….( each right angle )
Seg OB = seg OB
Δ OBC = Δ OBA ….( SAS test )
OC = OA
But seg OA is a radius.
Seg OC must also be radius.
C lies on the circle.
l
C
B
A
O
….( c.s.c.t )

8. That means line l intersects the circle in two distinct
points A and C.
l
C
B
A
O
But line l is a tangent. …….(Given)
It intersects the circle in only one point.
Our assumption that line l is not perpendicular to radius OA
is wrong.
Line l | radius OA.

9. THEOREM
P
A
l
If seg PA ^ line l
then line l is a tangent to the circle at point A.
A line perpendicular to the radius at its outer end
is a tangent to the circle.
M

10. THEOREM
O
(i) A circle with centre O.
(ii) P is a point in the exterior of the circle.
(iii) PA and PB are the tangents from P.
P
A
B
Result :
Given :
PA = PB
The lengths of two tangent segments drawn from an external
point to a circle are equal.

11. THEOREM :
The lengths of two tangent segments drawn from an external
point to a circle are equal.
O
Given : (i) A circle with centre O.
(ii) P is a point in the exterior of the circle.
(iii) PA and PB are the tangents from P. P
A
B
To prove : PA = PB
Construction : Draw OA, OB
Proof : In DPAO and DPBO,
mPAO = mPBO = 90o [Radius is perpendicular to the tangent]
Hypt. OP ≅ Hypt. OP [Common side]
II
seg OA ≅ seg OB [radii of the same circle]
DPAO ≅ DPBO [Hypotenuse side theorem]
seg PA ≅ seg PB [c.s.c.t]
and OP
PA = PB

12. Practice set 3.1
1. In the adjoining figure radius of a circle with centre C is 6 cm,
line AB is a tangent at A. Answer the following questions.
1) What is the measure of CAB ? Why ?
2) What is the distance of point C from line AB ? Why ?
3) d(A,B) = 6 cm, find d(B,C).
4) What is the measure of ABC ? Why ?
C
A B
Solution : Draw seg CA and seg CB.
i)CA | AB ( Tangent Theorem )
∴ ∠ CAB = 90 °
ii) Distance of point C from line AB is 6cm.
( Perpendicular distance)
iii)
CA ( radius) = 6 cm
d (A,B) = 6cm.
In Δ CAB, ∠ CAB = 90 °
∴ CB
2
+
=CA
2
AB
2
= 6
2
+ 6
2
= 36 + 36
CB
2
= 72
(Pythagoras theorem)
∴ CB = 6√2 (Taking sq. root)
iv) In Δ CAB,
CA = AB ( Each 6cm)
∴ ∠ ACB = ∠ ABC ( Isosceles triangle
theorem)
∠ ACB + ∠ ABC + ∠ CAB = 180
∠ ABC+ ∠ ABC + 90 = 180
2∠ ABC = 180 - 90
∴
2∠ ABC = 90
∠ ABC = 45 °
( i )
( ii )
[ From i
& ii ]

13. 2. In the adjoining figure, O is the centre of the circle.
From point R, seg RM and seg RN are tangent
segments touching the circles at M and N.
If (OR) = 10 cm and radius of the circle = 5 cm.
1) What is the length of each segment ?
2) What is the measure of MRO ?
3) What is the measure of MRN ?
O R
M
N
Solution : Draw seg OM and seg ON.
OM = ON = 5cm
i)OM | MR ( Tangent Theorem )
∴ ∠ OMR = 90 °
∴ OR
2
+
=OM
2
MR
2
= 5
2
+ MR
2
= 25 + MR
MR
2
= 100 - 25
By Pythagoras theorem,
10
2
100
2
MR = 75
2
∴ MR = 5√3 cm
MR = NR ( Tangent Segment Theorem )
∴ NR = 5√3 cm
ii) In Δ OMR,
Similarly, ∠ NRO = 30 °
iii)
∠MRN = ∠ MRO + ∠ NRO
[ Angle sum property ]
= 30 + 30 [ From ii ]
∴ ∠ MRN = 60 °
OM = OR [ Given ]
∴ ∠ MRO = 30 ° [ Converse of 30-60
-90 theorem ]

14. 3. Seg RM and seg RN are tangent segments of
a circle with centre O. Prove that seg OR bisects
MRN as well as MON.
O R
M
N
Solution : Seg RM and seg RN are tangents from
external point R.
∴ Seg RM = seg RN ( Tangent Segment Theorem )
…( i )
OM | MR ( Tangent Theorem )
∴ ∠ OMR = 90 ° …( ii )
ON | NR ( Tangent Theorem )
∴ ∠ ONR = 90 ° …( iii )
In Δ OMR and Δ ONR
Hypo OR = Hypo OR ( Common side )
∠ OMR = ∠ ONR [ From ii & iii ]
Seg RM = seg RN [ From i ]
∴ Δ OMR @ Δ ONR [ By Hypo- side test ]
∴ ∠ MRO = ∠ ORN and ∠ MOR = ∠ NOR [ By c.a.c.t ]
∴ Seg OR bisect ∠ MRN as well as ∠ MON

15. 4. What is the distance between two parallel tangents of a circle having radius
4.5 cm ? Justify your answer.
O
A m
l
B
Solution :O is the centre of a circle .
l and m are two parallel tangent of the circle.
Radius of a circle = 4.5cm .
Distance between two parallel tangents = AB.
AB = OA + OB
= 4.5 + 4.5
∴ AB = 9 cm
Thus, distance between the parallel tangents l and m is 9cm.

16. THEOREM
If two circles are touching circles then the common point
lies on the line joining their centres.
Circles touching EXTERNALLY Circles touching INTERNALLY
O A
T
T
O A
O – T – A O – A – T

17. Practice set 3.2
1. Two circles having radii 3.5 and 4.8 cm touch each
other internally. Find the distance between their
centers. O
A T
Solution :
Let two circles with centre O and A touch each other
Internally at point T.
∴ O – A – T ( Theorem of touching Circles )
∴ OT = OA + AT ( O – A – T )
If two circles touch internally, the distance between their radii
Is equal to difference of their radii.
O T = 4.8 cm & AT = 3.5 cm ( Given )
∴ OA = OT - AT
∴ OA = 4.8 – 3.5
∴ OA = 1.3 cm
∴ The Distance between their centers is 1.3 cm.

18. 2. Two circles having radii 5.5 and 4.2 cm
touch each other externally. Find the
distance between their centers.
O A
T
Solution :
Let two circles with centre O and A touch each other
externally at point T.
∴ O – T – A ( Theorem of touching Circles )
∴ OA = OT + AT ( O – T-A )
If two circles touch externally, the distance between their radii
Is equal to sum of their radii.
O T = 5.5 cm & AT = 4.2 cm ( Given )
∴ OA = 5.5 + 4.2
∴ OA = 9.7 cm
∴ The Distance between their centers is 9.7 cm.

19. 3. If radii of two circle are 4 cm and 2.8 cm. Draw figure of these circles touching
each other – (i) externally (ii) internally
O
A T
O A
T
4cm 2.8cm

20. 4. In figure, the circles with centres P and Q
touch each other at R. A line passing through
R meets the circles at A and B respectively.
Prove that –
1) Seg AP || seg BQ,
2) DAPR~ DRQB , and
3) Find RQB if PAR = 35 ⁰
P Q
A
B
R
Solution : i) By theorem of touching circle,
Point P, R and Q are collinear.
∴ ∠ARP @ ∠QRB ( Vertically opp. angles )
…( i )
Now, seg AP @ seg PR ( Radii of the same circle )
∴ ∠ARP @ ∠PAR ( Isosceles triangle theorem )
…( ii )
Similarly, seg RQ @ seg QB ( Radii of the same circle )
∴ ∠QRB @ ∠QBR ( Isosceles triangle theorem )
…( iii )
From (i), (ii) and (iii),
∠PAR @ ∠QBR …( iv )
∠PAB @ ∠QBA …( A-R-B )
∴ Seg AP || seg BQ …( Alternate angle test )

21. ii) In Δ APR and Δ RQB,
∠ARP @ ∠QRB … [ From (i) ]
∠PAR @ ∠QBR … [ From (iv) ]
∴ Δ APR ~ Δ RQB ….(AA)
iii) ∠PAR = 35 ⁰
∠PAR = ∠QBR … [ From (iv) ]
∴ ∠QBR = 35 ⁰
∠QBR = ∠QRB … [ From (iii) ]
∴ ∠QRB = 35 ⁰ ….(v)
iv) In Δ RQB,
∠QBR + ∠QRB + ∠ RQB = 180 ⁰
35 + 35 + ∠ RQB = 180
70 + ∠ RQB = 180
∴
∴
∴ ∠ RQB = 180 - 70
∴ ∠ RQB = 110 ⁰
P Q
A
B
R

22. 5. In figure, the circles with centers A and B
touch each other at E. Line l is a common
tangent which touches the circles at C and
D respectively. Find the length of seg CD if
the radii of the circles are 4 cm and 6 cm.
A B
E
l
C
D
4cm
6cm
Construction:
4cm
M
Draw seg AM | seg DB such that,
D-M-B.
Solution:
6cm
CA = AE = 4cm(radius of same circle)
DB = BE = 6cm (radius of same circle)
In □ CAMD,
∠ACD = 90 ⁰
∠MDC = 90 ⁰
(Tangent theorem)
∠AMD = 90 ⁰ (Construction)
∠CAM = 90 ⁰ (Remaining angle)
∴ □ CAMD is rectangle (By definition)
∴ CA = DM = 4cm (Opp. Sides of rectangle)
DB = DM + MB (D-M-B)
∴ 6 = 4 + MB
∴ MB = 2 cm
In Δ AMB, ∠AMB = 90⁰(Construction)
∴ AB
2
+
=BM
2
AM
2
= 2
2
+ AM
2
= 4 + AM
AM
2
= 100 - 4
By Pythagoras theorem,
10
2
100
2
AM = 96
2
∴ AM = 4√6 cm
But, AM = CD
(Opp. Sides of rectangle)
∴ CD = 4√6 cm
(AB = 4 + 6 = 10)

23. Arcs of a circle
A
B
X
Y
A secant divides a circle in two parts. Any of these two
parts and the common points of the circle and the
secant constitute an arc of the circle.
Minor Arc
A minor arc is an arc smaller than a semicircle. Arc AXB is a minor arc.
Major Arc
A major arc is an arc bigger than a semicircle. Arc AYB is a major arc.
Central angle
An angle whose vertex is the centre of a circle is
called Central angle.

24. Remember This !
1) Measure of minor arc is equal to the measure of its corresponding
central angle.
2) Measure of major arc = 360 ⁰ – measure of minor arc.
3) Measure of a semi-circular arc is 180 ⁰.
4) Measure of a complete circle is 360 ⁰.
5) Two arcs are congruent if their measure and radii are equal.

25. Theorem
The chords corresponding to congruent arcs of a circle ( or congruent circles ) are
Congruent.
B
A
P C
D
Q
E
Given :In a circle with centre B arc APC ≅ arc DQE
To prove : Chord AC ≅ Chord DE
Proof : In Δ ABC and Δ DBE,
Side AB ( Radii of same circle )
≅ Side DB
Side ≅ Side
BC EB ( Radii of same circle )
∠ ABC ≅ ∠ DBE ( Measure of congruent arcs )
∴ Δ ABC ≅ Δ DBE ( By SAS Test )
∴ Chord AC ≅ Chord DE ( By c.s.c.t )
Hence , proved the chords corresponding to congruent arcs of a circle are
congruent.

26. Theorem
Corresponding arcs of congruent chord of a circle ( or congruent circles ) are
Congruent.
B
A
P C
D
Q
E
Given :In a circle with centre B chord AC ≅ chord DE
To prove : Arc APC ≅ arc DQE
Proof : In Δ ABC and Δ DBE,
Side AB ( Radii of same circle )
≅ Side DB
Side BC ( Radii of same circle )
≅ Side BE
Side AC ( Given )
≅ Side DE
∴ Δ ABC ≅ Δ DBE ( By SSS Test )
∴ ∠ ABC ≅ ∠ DBE ( By c.a.c.t )
∴ Arc APC ≅ arc DQE (measure of minor arc is equals to
corresponding central angle. )
Hence , proved the corresponding arcs of congruent chord of a circle are
congruent.

27. Practice set 3.3
1. In figure, points G, D, E, F are concyclic points
of a circle with centre C. C
G
D
E
F
∠ ECF = 70 ⁰, m ( arc DGF ) = 200 ⁰
Find m (arc DE ) and m ( arc DEF ).
70⁰
200⁰
Solution:
i) ∠ECF = m( arc EF ) (Measure of central angle is equals to the measure
of corresponding arc.)
∴ m( arc EF ) = 70⁰ …( i )
m( arc DGE ) = m(arc DGF) + m(arc EF)
= 200 + 70
m(arc DGE) = 270⁰ …( ii )
m( arc DE) = 360 - m(arc DGE)
= 360 - 270
(Measure of complete circle is 360 ⁰)
∴ m( arc DE) = 90 ⁰
ii) m( arc DEF ) = m(arc DE) + m(arc EF)
= 90 + 70 …[ From iii & i ]
∴ m( arc DEF) = 160 ⁰
…( iii )

28. 2. In figure DQRS is an equilateral triangle. Prove that
1) arc RS ≅ arc QS ≅ arc QR
2) m (arc QRS ) = 240 ⁰
Q
R S
Solution:
i) DQRS is an equilateral triangle.
∴ Chord QR ≅ chord RS ≅ chord QS…( i )(sides of an
equilateral D)
Chord QR ≅ chord RS…[ from i ]
∴ arc QR ≅ arc RS…( ii )( arcs corresponding to congruent chord)
Chord RS ≅ chord QS…[ from i ]
∴ arc RS ≅ arc QS…( iii )( arcs corresponding to congruent chord)
∴ arc RS ≅ arc QS ≅ arc QR [ from ii & iii ]
ii) m(arc QR) + m(arc RS) + m(arc QS) = 360 ⁰(Measure of complete circle is 360 ⁰)
…( iv )
∴ m(arc QS) + m(arc QS) + m(arc QS) = 360
∴ 3 m(arc QS) = 360
∴ m(arc QS) = 120 ⁰
∴ m(arc QR) = m(arc RS) = m(arc QS) = 120 ⁰
m(arc QRS) = m(arc QR) + m(arc RS)
= 120 + 120
∴ m(arc QRS) = 240 ⁰
…[ from iv ]

29. 3. In figure chord AB ≅ chord CD, prove that,
arc AC ≅ arc BD
A
C B
D
Solution:
Chord AB ≅ chord CD ( Given )
∴ arc AB ≅ arc CD …( i )( arcs corresponding to congruent
chord)
Subtract (arc CB) from eq. (i)
(arc AB) – (arc CB) ≅ (arc BD) – (arc CB)
∴ arc AC ≅ arc BD

30. B
A
C
X
A
B
C
D
INTERCEPTED ARC
A
B
C
A
B
C
X
A
B
C
D
E
A
C
B
X
Intercepted arc : AC Intercepted arcs : AC & DC Intercepted arcs : AXC & AC
Intercepted arcs : AC & DE Intercepted arc : BC Intercepted arc : AC

31. B
A C
X
What type of angle is this
?
INSCRIBED ANGLE THEOREM
The measure of an inscribed angle is half of the measure of
its intercepted arc.
=
1
2
mABC m(arc AXC)

32. The measure of an inscribed angle is half of the measure of
its intercepted arc. A
B C
X
O
Given: In a circle with centre O,  BAC is inscribed in arc BAC.
Arc  BXC is intercepted by the angle.
To prove :  BAC = × m(arc BXC)
Construction : Draw ray AO. It intersects the circle at X.
Draw radius OC.
Proof : In DAOC ,
Side OA = side OC ( radii of the same circle)
∠OAC = ∠OCA
∴ ( Isosceles triangle theorem)
Let ∠OAC = ∠OCA = x ( i )
Now, ∠XOC = ∠OAC + ∠OCA ( Exterior angle)
But, ∠XOC is a central angle.
∴ m(arc XC) = 2x ( Definition of measure of an arc.)
( ii )
= x + x = 2x
From i and ii
 OAC =  XAC = × m(arc XC) ( iii )

33. A
B C
X
O
Similarly, drawing seg OB we can prove
 XAB = × m(arc XB) ( iv )
∴  XAC +  XAB = × m(arc XC) + × m(arc XB)
∴ BAC = × m(arc BXC)
[ From iii & iv ]

34. COROLLARY 1
A B
O
X
C
mACB =
1
2
m(arc AXB)
[Inscribed angle theorem]
mACB = 180
mACB = 90
o
1
2

An angle inscribed in a semicircle is a right angle.
90

35. [Inscribed angle
theorem]
Angles inscribed in the same arc are congruent.
COROLLARY 2
A
C
B
D
mABC =
1
2
m(arc AC)
mADC =
1
2
m(arc AC)
mABC = mADC
ABC @ ADC
Note : Both these angles have the
same intercepted arc

36. [Inscribed angle theorem]
[Inscribed angle theorem]
THEOREM
The opposite angles of a cyclic quadrilateral are supplementary.
A
B
C
D
Given : oABCD is a cyclic quadrilateral.
To prove :
Their sum is 180º
= 180
o
mBAD + mBCD = 180
o
mABC
Proof :
=
=
+ mADC
...(i)
...(ii)
Adding (i) and (ii), we get,
mABC + mADC = m(arc ADC) +
mABC + mADC = [m(arc ADC) + m(arc ABC)]
mABC + mADC =
mABC + mADC = 180
o
...(iii)
× 360
1
2
1
2
1
2
m(arc ABC)
1
2
mABC
mADC
m(arc ADC)
m(arc ABC)
1
2
1
2
[Angular measure of a circle]

37. mBAD + mBCD + mABC
In oABCD,
+ mADC =
mBAD + mBCD + 180 = 360
mBAD + mBCD = 180
o
360
o
THEOREM
The opposite angles of a cyclic quadrilateral are supplementary.
Given : oABCD is a cyclic quadrilateral.
To prove : = 180
o
mBAD + mBCD = 180
o
mABC
Proof :
+ mADC A
B
C
D
mABC + mADC = 180
o
...(iii)
[Sum of the measure of angles of a
quadrilateral is 360º
[From (iii)]

38. A
D
B
C
If A + C = 180o
then oABCD is cyclic.
Sum is 180º
A and C
B and D
Consider
A and C
If opposite angles of a quadrilateral are supplementary,
then the quadrilateral is cyclic.

39. A
B
C
D
E
An exterior angle of a cyclic quadrilateral
Here A, B, C, and
D are interior angles
of cyclic oABCD
An angle which forms a linear pair with an interior angle of
a quadrilateral is called an exterior angle of a cyclic quadrilateral.

40. A
B
C
D
E
COROLLARY
An exterior angle of a cyclic quadrilateral is congruent to
the angle opposite to its adjacent-interior angle
oABCD is a cyclic quadrilateral.
[The opposite angles of a cyclic
quadrilateral are supplementary]
mDCE + mDCB = 180º …(i)
mDCB + mBAD = 180º …(ii)
mDCE + mDCB = mDCB + mBAD
mDCE = mBAD
DCE @ BAD
DCB is an adjacent-interior
angle of exterior DCE
Exterior
angle
Adjacent-interior
angle

41. If two points on a given line subtend equal angles at two distinct points
which lie on the same side of the line, then the four points are
concyclic.
Theorem
B C
A D
Point B and C lie on the same side of the line AD.
∠ABD = ∠ACD
∴ Points A, B, C and D are concyclic.

42. Remember this!
A
D
E
C
B
i) ∠ABD = × [ m(arc AD) + m(arc EC) ]
ii) ∠DBC = × [ m(arc AE) + m(arc DC) ]
E
C
A
D
B
i) ∠BED = × [ m(arc BD) - m(arc AC) ]

43. Practice set 3.4
Q1. In figure a circle with centre O, length of
chord AB is equal to the radius of the circle.
Find the measure of each of the following.
1)  AOB 2)  ACB
3) Arc AB 4) arc ACB
O
A
B
C
Solution:Chord AB is equal to the radius.
i) In DAOB,
side AB = side OA = side OB (Given)
∴ DAOB is an equilateral triangle.
∴ ∠AOB = 60 ⁰ (Angle of equilateral triangle.)
…( i )
ii) m(arc AB) = 60⁰ …( ii )(Measure of central angle)
m∠ACB = × m(arc AB)
= × 60
m∠ACB = 30⁰
iii) m(arc AB) = 60⁰ [ From ii ]
iv) m(arc ACB) = 360 – m(arc AB)
= 360 – 60
m(arc ACB) = 300⁰

44. Q2. In figure □ PQRS is cyclic. Side PQ ≅ side
QR.  PSR = 110º, Find –
1)  PQR 2) m ( arc PQR)
3) M (arc QR) 4)  PRQ
S
R
Q
P
Solution:□ PQRS is cyclic .
i)m∠PQR + m∠PSR = 180 (Opp. Angles of cyclic ⧠)
∴ m∠PQR + 110 = 180
∴ m∠PQR =180 - 110
∴ m∠PQR = 70⁰ …( i )
ii) m∠PQR = × m(arc PSR) (Inscribed angle theorem)
70 = × m(arc PSR )
∴ m(arc PSR ) = 140⁰
m(arc PQR) = 360 - m(arc PSR)
= 360 - 140
∴ m(arc PQR) = 220⁰
…( ii )
…( iii )

45. Q2. In figure □ PQRS is cyclic. Side PQ ≅ side
QR.  PSR = 110º, Find –
1)  PQR 2) m ( arc PQR)
3) M (arc QR) 4)  PRQ S
R
Q
P
iii) m(arc PQ) = m(arc QR) ( Given & corresponding
Arcs of congruent chords.)
∴ m(arc PQR) = m(arc QR) + m(arc QP)
220 = m(arc QR) + m(arc QR)
220 = 2 m(arc QR)
∴ m(arc QR) = 110 ⁰
…( iv )
…( v )
iv) m∠PRQ = × m(arc PQ) (Inscribed angle theorem)
= × 110 [ From iv & v ]
∴ m∠PRQ = 55 ⁰

46. Q5. Prove that, any rectangle is a cyclic
quadrilateral.
A B
C
D
Given: □ ABCD is rectangle .
To prove: □ ABCD is cyclic .
Proof: □ ABCD is rectangle .
∴ m∠ABC = 90 ⁰
m∠BCD = 90 ⁰
m∠CDA = 90 ⁰
m∠DAB = 90 ⁰
(Angles of rectangle.)
…( i )
m∠ABC + m∠CDA = 180 ⁰ …( ii ) ( Opp. angles of rectangle.)
Similarly, m∠BCD + m∠DAB = 180 ⁰ …( iii )( Opp. angles of rectangle.)
∴ □ ABCD is cyclic . [ From ii & iii and theorem of converse of
cyclic quadrilateral

47. Q6. In figure, altitude YZ and XT of DWXY
intersect at P. Prove that,
1) □ WZPT is cyclic.
2) Point X, Z, T, Y are concyclic.
X
Z
W
Y
T
P
Given: Altitudes YZ and XT of D WXY intersect at P. .
To prove: i) □ WZPT is cyclic .
ii) Points X, Z, T, Y are concyclic.
Proof: i) In □ WZPT ,
m∠PZW= m∠PTW = 90⁰ (Given)
m∠PZW + m∠PTW = 180 ⁰ (Opp. Angles of cyclic quadrilateral)
Hence, we can prove □ WZPT is cyclic quadrilateral.
ii) m∠XZY= m∠YTX (Each 90⁰)
Point T and point Z lie on the same side of the line XY.
If two points on a given line subtend equal angles at two distinct points
which lie on the same side of the line, then the points are concyclic.
Hence, we can prove points X, Z, T, Y are concyclic.

48. Q7. In figure, m( arc NS ) = 125º, m( arc EF ) = 37º,
find the measure  NMS.
M
E
F
N
S
Solution:
m(arc NS) = 125 ⁰
m(arc EF) = 37 ⁰
m∠NMS = × [ m(arc NS) - m(arc EF)
= × [ 125 – 37 ]
= × 88
∴ m∠NMS = 44 ⁰

49. Q8. In figure, chord AC and DE intersect at B. If
 ABE = 108 º, m( arc AE ) = 95º, find m( arc DC ).
A
D
E
C
B
Solution:
m∠ABE= 108⁰
m(arc AE) = 95 ⁰
m∠ABE = m∠DBC =108⁰ (Vertically opposite angle.)
m∠DBC = × [ m(arc AE) + m(arc DC) ]
∴ 108 = × [ 95 + m(arc DC) ]
∴ 108 × 2 = 95 + m(arc DC)
∴ 216 = 95 + m(arc DC)
∴ 216 - 95 = m(arc DC)
∴ m(arc DC) = 121 ⁰

50. A secant divides the circular region into two parts.
Each part is called a segment of the circle.
B
A
R1
R2
SEGMENT OF A CIRCLE
ALTERNATE SEGMENT
Each of the two segments formed by the
secant of a circle is called alternate segment
in relation with the other.
R1 and R2 are the two segments of the
circle formed by secant AB.
R1 is the alternate segment of R2.
Vice-versa, R2 is the alternate segment of R1.

51. Theorem of internal division of chords
If two secants AB and CD of a circle intersect inside
the circle in a point P then AP × BP = CP × DP
Given : Secants AB and CD intersect in point P.
To prove : AP × BP = CP × DP
Construction : Draw AC and DB.
Proof :
In DAPC and DDPB
APC @ DPB [Vertically opposite angles]
CAP @ BDP [Angles inscribed in the same arc]
DAPC ~ DDPB [AA test]
AP
DP
=
CP
BP
[Corresponding sides of
similar triangles]
AP × BP = DP × CP
P
A
B
D
C
Hint :
AP
DP
=
CP
BP

52. [The exterior angle of a cyclic quadrilateral
is congruent to the angle opposite to its
adjacent-interior angle]
If two secants AB and CD of a circle intersect outside
the circle in a point P then AP × BP = CP × DP
Given : Secants AB and CD intersect in point P.
To prove : AP × BP = CP × DP
Construction : Draw AC and DB. P
A
B
D
C
Proof : In DAPC and DDPB
PAC @ PDB
APC @ DPB [Common angle]
DAPC ~ DDPB [AA test]
Hint :
AP
DP
=
CP
BP
AP
DP
=
CP
BP
[Corresponding sides
of similar triangles]
AP × BP = CP × DP
Theorem of external division of chords

53. Theorem of angle between tangent and secant
If an angle has its vertex on the circle, its one side touches the circle and
the other intersects the circle in one more point, then the measure of the
angle is half the measure of its intercepted arc.
A
M
B
D
C
M
B C
A
D
A
F
E B C
M
D
Given: Let  ABC be an angle, where vertex B lie on a circle with centre M.
Its side touches the circle at B and side BA intersects the circle at A.
Arc ADB is intercepted by  ABC.
To prove: m∠ABC = × m(arc ADB)

54. Case 1
Proof : In figure centre M lies on the arm BA of  ABC,
A
M
B
D
C
 ABC =  MBC = 90 ⁰ (tangent segment theorem)
…( i )
Arc ADB is semicircle.
m(arc ADB) = 180 (definition of measure of arc)
…( ii )
From I and II
m∠ABC = × m(arc ADB)

55. Case 2
M
B C
A
D
x
x
y
Proof : In figure centre M lies in the exterior of  ABC,
Draw radii MA and MB.
Now,  MBA =  MAB (isosceles triangle theorem)
 MBC = 90 ⁰ (tangent theorem)
…( i )
Let,  MBA =  MAB = x and  ABC = y
 AMB + x + x = 180
 MBC =  MBA +  ABC = x + y
x + y = 90
2x + 2y = 180
In, DAMB, 2x +  AMB = 180
2x +  AMB = 2x + 2y
 AMB = 2y
 AMB = y
m(arc ADB) =  ABC
 AMB + 2x = 180
(Angle sum property)
(Multiplying both side by 2)

56. Case 3
A
F
E B
M
D
Proof : In figure ray AB is opposite ray of ray BC.
C
m∠ABE = × m(arc AFB) (proved in ii )
180 -  ABC = × m(arc AFB)
=
× [ 360 - m(arc ADB) ]
180 -  ABC = 180 - m(arc ADB) ]
m∠ABC = × m(arc ADB)
180 -  ABC =
ABE +  ABC = 180
 ABE = 180 -  ABC

57. If PAB is a secant to a circle intersecting at points
A and B and PT is a tangent then PA × PB = PT².
T
P
A
B
Given : A secant through the point P
intersects the circle in points A and B.
Tangent drawn through the point P
touches the circle in point T.
To prove : PA × PB = PT2
Construction : Draw BT and AT.
In DPTA and DPBT
PTA @ PBT [Angles in alternate segments]
TPA @ BPT [common angle]
DPTA ~ DPBT [AA test]
PT
PB
=
PA
PT
[Corresponding sides of
similar triangles]
PA × PB = PT2
TANGENT-SECANT THEOREM
i.e. PA × PB = PT × PT
Hint :
PA
PT
=
PT
PB
PA
PT
=
PT
PB

58. Practice set 3.5
Q1. In figure, ray PQ touches the circle at point
Q. PQ = 12, PR = 8, find PS and RS.
S
R
P
Q
Solution:
PQ = 12, PR = 8
PQ 2 = PR × PS (Tangent secant segment theorem)
∴ 12 2 = 8 × PS
∴ 144 = 8 × PS
∴ PS = 18 units
RS = PS - RP
= 18 - 8
∴ RS = 10 units

59. Q2. In figure, chord MN and chord RS intersect
at point D.
1) If RD = 15, DS = 4, MD = 8 find DN
2) If RS = 18, MD = 9, DN = 8 find DS
M
R
N
S
D
Solution:
1) MD × DN = DS × RD (theorem of internal division of
chord)
∴ 8 × DN = 4 × 15
∴ 8 × DN = 60
∴ DN = 7.5 units
2) MD × DN = DS × RD
Let, DS = x and RD = 18 - x
9 × 8 = x × (18 – x)
∴ 72 = 18x – x2)
∴ x2 - 18x + 72 = 0
∴ x2 - 12x - 6x + 72 = 0
∴ x(x – 12) – 6(x - 12) = 0
∴ (x – 12) (x – 6) = 0
∴ (x – 12) =0 or (x – 6) = 0
∴ x = 12 or x = 6
∴ DS = 12 unit or DS = 6 unit

60. Q3. In figure, O is the centre of the circle and B
is a point of contact. Seg OE | seg AD, AB = 12,
AC = 8, find i) AD ii) DC iii) DE
D
E
C
A
B
O
Solution:
1) AB = 12, AC = 8
AB 2 = AC × AD (Tangent secant segment theorem)
∴ 12 2 = 8 × AD
∴ 144 = 8 × AD
∴ AD = 18 unit
2) DC = AD - AC
= 18 - 8
∴ DC = 10 unit
3) DE = × DC (Perpendicular drawn from the centre of circle to chord
bisects the chord.)
= × 10
∴ DE = 5 unit

61. Q4. In figure, if PQ = 6, QR = 10, PS = 8 find TS.
T
R
S
Q
P
Solution:
PT × PS = PR × PQ (theorem of external division of
chord)
PR = PQ + QR
= 6 + 10
∴ PR = 16 unit ( i )
∴ PT × PS = PR × PQ
PT × 8 = 16 × 6
∴
∴ PT × 8 = 96
[ From i ]
∴ PT = 12 unit
TS = PT - PS
∴ TS = 12 - 8
∴ TS = 4 unit

62. Q5. In figure, seg Ef is a diameter and seg DF is
a tangent segment. The radius of the circle is r.
Prove that, DE × GE = 4r2
E F
H
G
D
Solution:
Radius of the circle is r.
∴ EH = HF = r
DF 2 = DG × DE (Tangent secant segment theorem)
In DDFE,  DFE = 90
DE 2 = DF2 + EF2
( i )
DF 2 = DE 2 – EF 2 ( ii )
Substituting eq. ii in eq. i
DE2- EF2 = DG × DE
DE2 – (2r)2 = DG × DE
DE2 – 4r2 = DG × DE
DE2 - DG × DE = 4r2
DE ( DE – DG) = 4r2
DE × GE = 4r2
( DE – GD = GE )

63. Problem set 3
Q9. In figure, line l touches the circle with
centre O at point P. Q is mid point of radius
OP. RS is a chord through Q such that chords
RS || line l. If RS = 12 find the radius of the
circle.
Solution:
 OPC = 90
R
O
Q
P
S
l
C
(Tangent theorem)
( i )
Line l || Chord RS, on transversal OP,
 OPC =  OQR ( Converse of corres-
ponding angle test)
( ii )
∴  OQR = 90 ( iii ) [ From i & ii ]
∴ Seg OQ | chord RS at point P
∴ QR = × RS
∴ QR = × 12
∴ QR = 6
Let the radius of circle be 2x
∴ OR = OP = 2x
∴ OQ = × OP
∴ OQ = × 2x
∴ OQ = x
In DOQR,  OQR = 90
∴ OR2 = OQ2 + RQ2
∴ 2x2 = x2 + 62
∴ 4x2 - x2 = 36
∴ 3x2 = 36
∴ x2 = 12
∴ x = 2√3
∴ OR = 2√3 × 2
∴ OR = 4√3

64. Q22. In figure, two circles intersect each
other at points S and R. Their common
tangent PQ touches the circle at point P,
Q. Prove that,  PRQ +  PSQ = 180
R
P Q
S
Solution:
 RSP =  RPQ ( i )
 RSQ =  RQP( ii )
(Angles in alternate segment)
In D RPQ,
 RPQ +  RQP +  PRQ = 180
 RSP +  RSQ +  PRQ = 180 [ From i & ii ]
 PSQ +  PRQ = 180 [ Angle addition property ]
∴

65. Q23. In figure, two circles intersect each
other at points M and N. Secants drawn
through M and N intersect the circles at
points R, S and P, Q respectively.
Prove that seg SQ || seg RP
P N
Q
R
M
S
Construction: Draw seg MN
Solution: □ MNRP is cyclic. [ By definition]
 PRM +  PNM = 180
∴ ( i ) [ Opp. Angles of cyclic □.]
□ MNQS is cyclic. [ By definition]
 PNM =  MSQ
∴ ( ii )[ Exterior angle of a cyclic □ is equal to
its interior opposite angle.]
 PRM +  MSQ = 180
∴ [ From i & ii ]
 PRS +  RSQ = 180
∴ [ R-M-S ]
This is a pair of interior angles on transversal RS.
∴ Seg SQ || seg RP [ By interior angle test ]

66. Q5. In figure, □ ABCD is a parallelogram.
IT circumscribes the circle with centre T.
Point E, F, G, H are touching points. IF
AE = 4.5, EB = 5.5, find AD.
A
H
D
E
G
F
T
B
C
Solution:
AE = AH = 4.5
EB = BF = 5.5
Let,
HD = DG = x
GC = CF = y
[ Tangent segment theorem]
□ ABCD is a parallelogram.
AB = DC [ Sides of a parallelogram]
∴ AE + EB = DG + GC
∴ 4.5 + 5.5 = x + y
∴ x + y = 10
AD = BC [ Sides of a parallelogram]
∴ AH + HD = BF + FC
∴ 4.5 + x = 5.5 + y
∴ x – y = 5.5 – 4.5
∴ x – y = 1
[ i ]
[ ii ]
Adding i & ii, we get
x + y + x – y = 10 + 1
∴ 2x = 11
∴ x = 5.5
∴ HD = 5.5
AD = AH + HD
∴ AD = 4.5 + 5.5
∴ AD = 10 unit

67. Q4. In figure, O is the centre of the circle.
Seg AB, seg AC are tangent segments.
Radius od the circle is r and l(AB) = r,
Prove that, □ ABOC is a square.
Solution: Draw OB and OC.
A
B
C
O
r r
r
In □ ABOC,
Seg AB = seg AC = r (i) [ Tangent segment theorem]
Seg BO = seg OC = r (ii) [ Radii of the same circle ]
∴ BO = OC = AB = AC [ From i & ii ]
Hence, □ ABOC is rhombus [ By definition ]
 OBA =  OCA = 90 ( iii ) [ Tangent theorem ]
AC || BO [ Opp. Sides of rhombus are parallel ]
∴  OBA +  CAB = 180 [ Interior angles ]
∴ 90 +  CAB = 180
∴  CAB = 90 (iv)
Similarly,  BOC = 90 (v)
 OBA =  OCA =  CAB =  BOC = 90 [ From (iii) , (iv) & (v) ]
Hence, □ ABOC is square. [ By definition ]

68. Q10. In adjacent figure segment AB is a
diameter of a circle with Centre C. Line
PQ is a tangent which touches the circle
at point P. Seg AP | line PQ and seg PQ |
line PQ. Prove that seg CP = seg CQ
Construction : Draw seg CP, seg CT, seg CQ.
Solution:
Seg CT | PQ (i) [ Tangent theorem]
∴  CTP =  CTQ = 90 [ ii ]
Seg AP || seg CT || seg BQ [ Given & from( i )]
∴ = [ Property of three parallel lines ]
Seg AB = seg CB [ Radii of the same circle ]
∴ = 1
∴ Seg PT = seg TQ [ iii ]
In DCTP and DCTQ
Seg CT = seg CT (Common side)
 CTP =  CTQ [From (i)]
Seg PT = seg TQ [From (iii)]
∴ DCTP ≅ DCTQ [SAS test]
∴ Seg CP = seg CQ [c.s.c.t]

69. Q8. In adjacent figure, circles with
centers X and Y touch internally at point
Z. Segment BZ is a chord of bigger circle
and it intersects smaller circle at point A.
Prove that, seg AX || seg BY.
Construction : Draw seg YZ.
Solution: In smaller circle,
Seg AX = seg XZ [ Radii of the same circle ]
 XAZ =  AZX [ isosceles triangle theorem ]
 XAZ =  BZY (i) [ Z-X-Y, Z-A-B ]
In bigger circle,
Seg YZ = seg BY [ Radii of the same circle ]
 YBZ =  BZY (ii) [ isosceles triangle theorem ]
∴  XAZ =  YBZ [ From (i) and (ii) ]
∴ Seg AX || seg BY [ Converse of corresponding angle test ]

70. Q19. In figure, circles with centre C and D
touch internally at point E. D lies on the
inner circle. Chord EB of the outer circle
intersects inner circle at point A. Prove
that, seg EA ≅ seg AB.
Solution: Draw seg AD, BX and EX.
 EAD = 90 (i) [ Angle inscribed in semicircle ]
ED is diameter a of circle with centre C.
∴
∴ Seg AD | EB
∴ Seg EA ≅ seg AB(Perpendicular drawn from the centre of circle to chord
bisects the chord.)
Hence , Proved Seg EA ≅ seg AB.
EB is a chord of circle with centre D.
E
A
B
D
C