1. CHAPTER 10

2. SOME DEFINITIONS
• CIRCLES- A circle is a collection of those points in a plane
that are at a given constant distance from a given fixed point in
the plane.
r
O X

3. The collection of all points lying inside and on the
circle C(O , r) is called a circular disc with centre O
and radius r.
Circles having the same centre
but with different radii are said
to be concentric circles. . O

4. A continuous piece of a circle is
called an arc of the circle.
r
O.
P6
P2
P4
P5
P1
P3
Consider a circle C(O , r). Let
P1, P2, P3, P4, P5, P6 be points
on the circle. Then, the pieces
P1,P2,P3,P4,P5,P6 etc. are all
arcs of the circle C(O, r)

5. Arc
Minor Arc – A minor arc of a circle is
the collection of those points that lie on
and also inside a central angle.
Major Arc - A major arc of a circle is
the collection of points of the circle
that lie on or outside a central angle

6. CHORD :-
A line segment
joining any two
points on a circle
is called a chord
of the circle.
A chord passing
through the centre of
a circle is known as
its diameter.
P
Q
r rO
P Q
QP
PQ
d

7. THEOREM’S
Given- Arc PQ of a circle C(O , r) and arc RS of another circle C(O’ , r)
congruent to C(O , r) such that PQ ≅ RS.
To Prove- PQ=RS
Construction- Draw line segments OP, OQ, O’R and O’S
If two arcs of a circle are congruent ,
then corresponding chords are equal.
O
P Q
O’
R S

8. Proof :-
In triangles OPQ and O’RS , we have
OP=OQ=O’R=O’s=r [Equal radii of two circles]
∠ POQ = ∠ RO’S [∵ PQ ≅ RS m(PQ)= m(RS) ∠POQ=
∠RO’S]
So, by SAS criterion of congruence , we have
▲POQ ≅ ▲RO’S
PQ=RS
If PQ , RS are major arcs , then QP and SR are
minor arcs.
So, PQ ≅ RS , PQ=RS
QP=SR , QP=SR
Hence, PQ ≅ RS  PQ=RS

9. Given- A chord PQ of a circle C(O, r) and perpendicular OL to
the chord PQ.
To Prove- LP=LQ
Construction- Join OP and OQ.
O
P Q
L
Proof- In triangles PLO and QLO , we have
OP=OQ=r [Radii of the same
circle]
OL= OL
[Common]
And, ∠OLP= ∠OLQ [Each
equal to 90°]

10. THEOREM 3 :-
Given - A chord PQ of a circle C(O, r) with mid-point M.
To prove – OM |PQ
Construction - Join OP and OQ
Proof – In triangles OPM and OQM, we have
OP=OQ [Radii of the same circle]
PM=MQ [M is mid-point of PQ]
OM=OM
So, SSS-criterion of congruence , we have
▲OPM ≅ ▲OQM
 ∠OMP= ∠OMQ
But, ∠OMP+ ∠OMQ=180º
 ∠OMP+ ∠OMP=180º
 2∠OMP=180º
 ∠OMP=90º
 ∠OMP=∠OMQ=90º [∵ ∠OMP= ∠OMQ]
Hence, OM |PQ
O
P Q
M

11. Given- Three non-collinear points P,Q and R.
To Prove- There is one and only one circle
passing through P, Q and R.
Construction- Join PQ and
QR. Draw perpendicular
bisector AL and BM of PQ and
RQ respectively. Since P,Q,R are
not collinear. Therefore, the
perpendicular bisectors AL and
BM are not parallel. Let AL and
BM intersect at O. Join OP,OQ
and OR.
R
A
B M
QP
L
O

12. Proof – Since O lies on the perpendicular bisector of PQ.
Therefore,
OP=OQ , Again, O lies on the perpendicular bisector of
QR.
Therefore, OQ=OR
Thus OP=OQ=OR=r .
Taking O as the centre draw a circle of radius s. Clearly, C(O, s)
passes through P, Q and R. This proves that there is a circle
passing through the point P, Q and R.
We shall now prove that this is the only circle passing through P, Q
and R.
Let there be another circle with centre O’ and radius r, passing
through the points P,Q and R. Then, O’ will lie on the perpendicular
bisectors AL of PQ and BM of QR.
Since two lines cannot intersect at more than one point , so O’ must
coincide with O. Since OP=r, O’P=s and O and O’ coincide, we must
have
r=s
 C(O, r) ≅ C(O, s)
 Hence, there is one and only one circle passing through three non-
collinear points P, Q and R.

14. Ans 1: The below given steps will be followed to find the centre of
the given circle.
Step1. Take the given circle.
Step2. Take any two different chords AB and CD of this circle
and draw perpendicular bisectors of these chords.
Step3. Let these perpendicular bisectors meet at point O. Hence,
O is the centre of the given circle.

15. Ans 2: Consider two circles centered at point O and O’, intersecting
each other at point A and B respectively. Join AB. AB is the
chord of the circle centered at O. Therefore,
perpendicularbisector of AB will pass through O.
Again, AB is also the chord of the circle centered at O’. Therefore,
perpendicular bisector of AB will also pass through O’.
Clearly, the centers of these circles lie on the
perpendicular bisector of the common chord.

16. Q 3: Draw different pairs of circles. How many points
does each pair have in common? What is the
maximum number of common points?
Ans 3:
Consider the following pair of circles.
The above circles do not intersect each other at
any point. Therefore, they do not have any
point in common.

17. The below circles touch each other at 1
point X only. Therefore, the circles have
1 point in common.

18. These circles intersect each other at two points G and
H. Therefore, the circles have two points in common.
It can be observed that there can be a maximum of 2
points in common. Consider the situation in which
two congruent circles are superimposed on each
other. This situation can be referred to as if we are
drawing the circle two times.