This document provides information about circles and some circle theorems. It begins with definitions of basic circle terminology like radius, diameter, chord, arc, and sector. It then presents five circle theorems without proofs: 1) If two arcs are congruent, their chords are equal; 2) The perpendicular bisector of a chord passes through the center; 3) The line from the center to the midpoint of a chord is perpendicular to the chord; 4) There is only one circle passing through three non-collinear points; 5) The angle subtended by an arc at the center is double the angle at any other point on the remaining part of the circle. Sketch proofs are provided for theorems 2 through 5

1. Maths
presentation
circle
s

2. MOHIT PATIL
MAUSAM
MANISH
KIRTI
MAYURI
NAJIB
KEWAL
MAYANK

3. CONTENTS
Introduction
Arc of circle
 chord diameter of circle
Area of circle
Circles in real life application

4. INTRODUCTION
• CIRCLE
• A circle is a collection of all points in a plane that arc at a given fixed
point in th The distance between any of the points and the centre is
called the radius. It can also be defined as the locus of a point
equidistant from a fixed point. e plane
• . A circle is a simple closed curve which divides the plane into two
regions: an interior and an exterior. In everyday use, the term "circle"
may be used interchangeably to refer to either the boundary of the
figure, or to the whole figure including its interior; in strict technical
usage, the circle is the former and the latter is called a disk.

5. Arc of circle
A continuous piece of circle is called arc of a
circle
Minor arc
A collection of those points of the circle
That lie on and also inside a central angle
Major arc
major arc of a circle is the collection of points
of the circle that lie on and outside a central
angle

6. Chord, diameter,semicircle
A line segment joining any two points on a
circle is called a chord of circle
A chord passing through centre of circle is
known as diameter of circle
A diameter of circle divides divides it into two
equal parts which are arcs each of this two
arcs is called semicircle

7. Area of circle
• The ratio of a circle's circumference to its
diameter is π (pi), an irrational constant
approximately equal to 3.141592654. Thus
the length of the circumference C is related
to the radius r and diameter d by:

8. Circles in real life application
This is a clock. It has different shapes. Your clock can be a
square- shape. But have you notice what’s inside? I have
once open a destroyed one. It has a gear inside. Do you
know what a gear is? What is its shape? Yes it’s like a
circle. Have you seen a gear that has a shape other than
circle? Difinitely, you can not see a gear that has a shape
other than circle

9. Theorems
of circle

10. THEOREM 1
If two arcs of circle are congruent then
corresponding chords are equal

11. Sol:
• Given,
• Arc pq =arc ab and also congruent
• TO PROVE :- PQ=AB
• PROOF
• In OPQ andOAB
• OP=OQ=O’R=O’B (Equal radii of two circles )
• POQ= RO’S(ARC PQ =arc AB =>m{arc PQ }=m {AB}
• So by sas criterion of congruence we have POQ CONGRUENT
• AOB SO PQ=AB {C.P.C.T}
• HENCE PROVED

12. THEOREM 2
The perpendicular from center
of circle bisect the chord

13. SOI:
• Given
• A chord PQ of a circle
• Cons = join OP to OQ
• Proof in triangle PLO and QLO we have
• OP =OQ=r {radii of same circle}
• OL=OL (commom side )
• OLP= OLQ (each angle right
angle)
• So by RHS criterion of congruence we
have PLO= QLO

14. THE LINE JOINING THE
CENTRE OF CIRCLE TO THE
MID POINT OF CHORD IS
PERPEN DI CULAR TO
CHORD

15. SOL:-
• Given:- A chord PQ of a circle with mid point m
• Cons:-join OP to OQ
• PROOF :- in triangle OPM and OQM
• OP=OQ =r (radii of same circle)
• PM =MQ (M is mid point )
• OM=OM (COMMON SIDE}
• OPM= OQM
• OMP= OMP (C.P.C.T)
• OMP +OMQ=180 (LINEAR PAIR )
• 2 OMP =180
• OMP = 90
• OMP=OMQ =90
• OM PQ

16. THERE ARE ONE AND ONLY
ONE CIRCLE PASSING
THROUGH THREE NON CO
LINEAR POINTS
J
K

17. SOL :
• GIVEN ,three non collinear point P,Q,R AND O IS CENTRE OF
CIRCLE
TO PROVE THERE IS ONLY ONE CIRCLE PASS THROUGH THREE
NON COLLINEAR POINT
CONS= JOIN P TO Q TO R DRAW PERPENDICULAR BISECTOR AL
AND BM OF PQ AND RQ ALSO WHERE BOTH PERPENDICULAR
BISECTOR MEET NAME IT O;JOIN OR,OQ,OP
PROOF = IN TRIANGLE POJ AND QOJ
OJ=OJ (COMMON SIDE)
OJP=OJQ (EACH 90 )
PJ=JQ (PERPENDICULAR BISECTOR BISECT SIDE )
SO TRIANGLE OPJ = OJQ BY SAS CONGRUENCE SO
OP=OQ (C.P.C.T EQ 1 )
SIMARLY IN OQK AND OKR WE HAVE
OQ =OR ( C.P.C.T) EQ2
NOW FROM EQ 1 AND 2 WE GET
OP=OQ=OR

18. IF RADIUS ARE EQUAL
• SO THEN THE CIRCLE WILL PASS
THROUGH THREE NON COLLINEAR
POINTS
• HENCE PROVED

19. THE ANGLE SUBTENDED BY
ARC OF A CIRCLE AT
CENTRE IS DOUBLE THE
ANGLE SUBTENDED BY IT AT
ANY POINT ON THE
REMAINING PART OF CIRCLE

20. SOL=
GIVEN , AN MINOR PQ OF A CIRCLE AND
A POINT R ON REMAINING PART OF
CIRCLE i.e ARC QP
TO PROVE POQ=2 PRQ
CONS=JOIN RO PRODUCE
PROOF =
IN TRIANGLE ROP
PO=RO (RADII OF CIRCLE )
SO POR IS ISOSCELES TRIANGLE
SO RPO =PRO =x (IF SIDES ARE
EQUAL ANGLES ARE EQUAL )

21. • OQ= OR (RADII OF CIRCLE )
• SO ROQ IS ISOSCELES TRIANGLE
• ORQ = OQR (IF SIDES ARE
EQUAL ANGLES ARE EQUAL)
• POM IS EXTERIOR ANGLE OF
RPO SO RPO + PRO = POM
• x+x = POM (exterior angle of
triangle is equal to sum of two interior
angles) 2x = POM( EQ1)
In ROQ

22. Q
• NOW IN POQ
• ORQ +RQO = QOM (exterior angle
of triangle is equal to sum of two interior
angles) 2Y =QOM(EQ 2)
• FROM EQ 1 AND2 WE GET
• 2X+2Y =POM +QOM
• 2(X+Y)= POQ
• PRQ=1/2 POQ
• HENCE PROVED

23. THANK YOU