See the best explanation of class 9 circle ls. with theorems and easy proof understanding explaination very systematically organized

1. NAME – SIDDHI UDAY PAWAR.

2. WHAT IS CIRCLE?The collection of all the points in a plane, which are at a fixed
distance from a fixed point in the plane, is called a circle.
The fixed point is called the centre of the circle and the fixed
distance is called the radius of the circle.
The chord, which passes through the centre of the circle, is called a
diameter of the circle.
A piece of a circle between two points is called an arc. [Major arc and
minor arc]
The length of the complete circle is called its circumference.
The region between a chord and either of its arcs is called a
segment. [ Major segment and minor segment ]
The region between an arc and the two radii, joining the centre to

3. THEORE
MS ON
CIRCLES

4. THEOREM 1Equal chords of a circle subtend equal angles at the
centre.
Given- A circle C(O,r) & its 2 equal chords AB and CD.
RTP- angle AOB = angle COD
Proof- In ∆AOB & ∆COD, we have,
AB = CD [Given]
OA = OC [Each equal to r]
OB = OD [Each equal to r]
So, by SSS criterion of congruence, we have
∆AOB is congruent to ∆COD
=> Angle AOB = angle COD.
0
A
B C
D
Fig. 1

5. THEOREM 2
If the angles subtended by the chords of a circle at the
centre are equal, then the chords are equal.
Given- angle AOB = angle COD.
RTP- AB = CD
Proof- In the ∆s AOB & COD, we have
OA = OC [Each equal to r]
angle AOB = angle COD [Given]
OB = OD [Each equal to r]
So, by SAS criterion of congruence, we have
∆AOB is congruent to ∆COD
=> AB = CD.
0
A D
B C
Fig. 2

6. THEOREM 3The perpendicular from the centre of a circle to a chord
bisects the chord.
Given- OL is perpendicular to PQ
RTP- PL = LQ
Construction- Join OP & OQ
Proof- In ∆s PLO & QLO, we have
OP = OQ [Radii of the same circle]
OL = OL [Common]
angle OLP = angle OLQ [Each equal to 90º]
So, by RHS criterion of congruence, we have
∆PLO = ∆QLO
=> PL = LQ.
O
L
QP
Fig. 3

7. THEOREM 4
The line drawn through the centre of a circle to bisect a chord is
perpendicular to the chord.
Given- M is mid-point of PQ
RTP- OM is perpendicular to PQ
Construction- Join OP & OQ
Proof- In ∆s OPM & OQM, we have
OP = OQ [Radii of the same circle]
PM = MQ [M is the mid-point of PQ]
OM = OM [Common]
So, by SSS criterion of congruency, we have
∆OPM = ∆OQM
angle OMP = angle OMQ
But, angle OMP + angle OMQ = 180º
2 (angle OMP) = 180º
angle OMP = 90º So, OM is perpendicular to PQ.
M
P Q
O
Fig. 4

8. THEOREM 5
Q
There is one and only one circle passing through
three given non-collinear points.
Given- three non-collinear points P, Q & R.
RTP- there is only one circle passing through P, Q and R.
Construction-
1. Find a centre between angle PQR.
2. Taking the point O obtained as the centre & three
points of the circle draw a circle passing through them.
3. Thus, only one circle can pass through three points.
P
Fig. 5
R
O
A
L
M
B

9. THEOREM 6Equal chords of a circle ( or of congruent circles) are
equidistant from the centre ( or centres ).
Given- AB=CD
RTP- OL=O’M
Construction- Join OA & O’C.
Proof- Since, the perpendicular from the
centre of a circle to a chord bisects the chord.
Therefore, AL = BL
AL = ½ AB ………………………(i)
& CM = DM
O
A L B
O’
C M D
Fig. 6(i)
Fig. 6(ii)

10. THEOREM 6
Proof (continued)-
CM = ½ CD ……………………………..(ii)
Now, AB = CD [given]
½ AB = ½ CD
AL = CM …………………………………(iii)
Now, in ∆OAL & ∆O’CM,
OA = O’C [each equal to r]
angle OLA = angle O’MC [each 90º]
AL = CM
So, by RHS criteria, ∆OAL is congruent to ∆O’CM.
OL = O’M [CPCT]
O
A L B
O’
C
M
D
Fig. 6(i)
Fig. 6(ii)

11. THEOREM 7Chords equidistant from the centre of a circle are equal in length.
Given- OL = OM, OL & OM is perpendicular to AB & CD
RTP- AB = CD
Construction- Join OA & OC
Proof- Perpendicular from the centre of the of the circle bisect the chord.
AL = BL & AL = ½ AB
CM = DM & CM = ½ CD
In ∆ OAL & ∆OCM,
OA = OC [radius of the same circle]
angle OLA = angle OMC
OL = OM
So, by RHS criteria, ∆OLA is congruent to ∆OCM.
AL = CM [cpct]
½ AB = ½ CD
AB = CD.
C
A B
L
M
O D
Fig. 7

12. THEOREM 8
The angle subtended by an arc at the centre is the double
the angle subtended by it at any point on the remaining
part of the circle.
RTP- angle POQ = 2 (angle PRQ)
Construction- Join RO to M.
Proof- An exterior angle of a ∆ is equal to the sum of the interior
opposite angles.
In ∆POR, angle POM = angle OPR + angle ORP
angle POM = angle ORP + angle ORP [OP =OR = r]
angle POM = 2 (angle ORP) …………………..(i)
Similarly in ∆QOR, angle QOM = 2 (angle ORQ) …………………..(ii)
Adding (i) & (ii), angle POQ = 2 (angle PRQ).
M
P
O
R
Q
Fig. 8

13. THEOREM 9Angles in the same segment of a circle are equal.
RTP- angle PRQ = angle PSQ
Construction- Join OP & OQ
Proof- The angle subtended by an arc at the centre is double the
angle subtended by the arc at any at the part of the circle.
In fig. 9(i), angle POQ = 2 (angle PRQ) & angle POQ = 2 (angle PSQ)
2 (angle PRQ) = 2 (angle PSQ)
angle PRQ = angle PSQ.
In fig. 9(ii), reflex angle POQ = 2 (angle PRQ)
& reflex angle POQ = 2 (angle PSQ)
angle PRQ = angle PSQ
In both the cases we have, angle PRQ = angle PSQ.
Fig. 9(i)
Fig. 9(ii)
P
O
Q
R
S
P
O
Q
R
S

14. THEOREM 10
If a line segment joining two points subtends equal angles at
two other points lying on the same side of the line segment,
the four points lie on a circle (i.e. they are concyclic).
Given- angle PRQ = angle PSQ
RTP- P, Q, R, S are concyclic
Construction- Draw circle through P, Q, R.
Proof- If possible let S not be lying on this circle. Then this circle will
intersect at PS produced S’. Join S’Q.
Since, R & S’ are on the same side of PQ & the angles in the same segment
are equal.
angle PRQ = angle PS’Q = angle PSQ [ angle PRQ = angle PSQ]
So, S’ coincides with S. So, P, Q, R, S lies on the same circle.
P Q
R
S
S’
Fig. 10

15. THEOREM 11The sum of either pair of opposite angles of cyclic quadrilateral
is 180º.
RTP- angle A + angle C = 180º & angle B + angle D = 180º
Proof- angle ACB & angle ADB are angles in the same segment determined by
chord AB of the circle. So, angle ACB = angle ADB. …………………….(i)
Similarly, by chord BC, angle BAC = angle BDC ………………(ii)
Adding (i) and (ii), angle ACB + angle BAC = angle ADB + angle BDC.
angle ACB + angle BAC = angle ADC
angle ABC + angle ACB + angle BAC = angle ABC + angle ADC
180º = angle ABC + angle ADC
angle B + angle D = 180º
Again, angle A + angle B + angle C + angle D = 360º
angle A + angle C = 360º - [angle B + angle D]
angle A + angle C = 180º
C
B
A
D
Fig. 11

16. THEOREM 12
If the sum of a pair of opposite angles of a quadrilateral is
180º, the quadrilateral is cyclic.
Given- angle B + angle D = 180º
RTP- ABCD is a cyclic quadrilateral
Proof- If possible, let ABCD not a cyclic quadrilateral. Draw a circle
passing through A, B & C. suppose the circle meets AD’ extended to D.
Join D’C.
Now, ABCD’ is a cyclic quadrilateral.
Thus, angle ABC + angle AD’C = 180º ……………..(i)
But, angle B + angle D = 180º [given]
i.e, angle ABC + angle ADC = 180º ………………(ii)
So, angle AD’C = angle ADC.
So, D’ & D coincides. Hence, ABCD is cyclic quadrilateral.
A
BC
D
’D
Fig. 12