2. A Circle features…….
the distance around the Circle…
… its PERIMETER
Diameter
Radius … the distance across the
circle, passing through the centre
of the circle
Radius
Diameter...
the distance from the centre of
the circle to any point on the
circumference
3. A Circle features…….
Chord -> a line joining two points
on the Circumference.
… chord divides circle into
two segments
ARC -> part of the
circumference of a circle
Major
Segment
Minor
Segment
Tangent -> a line which touches
the circumference at one point
only
4. A perpendicular from the centre of a circle to
a chord bisects the chord.
Given: AB is a chord in a circle with centre O. OC ⊥ AB.
To prove: The point C bisects the chord AB.
Construction: Join OA and OB
Proof: In ∆ OAC and ∆ OBC,
∠OCA = ∠OCB = 90…………(Given)
OA = OB …………………..(Radii)
OC = OC ……………….(common side)
∠OAC = ∠OBC ………………………..(RHS)
CA = CB (corresponding sides)
The point C bisects the chord AB.
Hence the theorem is proved.
0
A B
O
C
5. The line drawn through the centre of a circle to bisect
a chord is perpendicular to the chord.
Given: AB be a chord of a circle with centre O and O is joined to the mid-
point M of AB.
To prove: OM ⊥ AB
Construction:
Join OA and OB.
Proof:
In ∆ OAM & ∆ OBM,
OA = OB……………. (Radii of a circle)
AM = BM ……………..(Given)
OM = OM (Common)
Therefore, ∆OAM ≅ ∆OBM ……………………….(SSS Rule)
This gives ∠OMA = ∠OMB = 90°
0
A B
O
M
6. Equal chords of a circle subtend equal
angles at the centre.
Given: Two equal chords AB and CD of a circle with centre O
To Prove: ∠ AOB = ∠ COD
In ∆ AOB and ∆ COD,
OA = OC……. (Radii of a circle)
OB = OD………..(Radii of a circle)
AB = CD ………...(Given)
Therefore,
∆ AOB ≅ ∆ COD…………. (SSS rule)
This gives ∠ AOB = ∠ COD
(Corresponding parts of congruent triangles)
OA
B
C
D
7. The angle subtended by an arc at the centre is double the
angle subtended by it at any point on the remaining part of
the circle.
Given: An arc PQ of a circle subtending angles POQ at the
centre O and PAQ at a point A on the remaining part of the
circle.
To prove: ∠ POQ = 2 ∠ PAQ.
P
Q
O
O O
P
Q
A
Q
AA
P
(i) arc PQ is minor (ii)arc PQ is a semicircle iii)arc PQ is major.
8. Construction: Let us begin by joining AO
and extending it to a point B.
Proof: In all the cases,
∠ BOQ = ∠ OAQ + ∠ AQO
because an exterior angle of a triangle is
equal to the sum of the two interior opposite angles
Also in ∆ OAQ,
OA = OQ …………………… (Radii of a
circle)
Therefore, ∠ OAQ = ∠ OQA (Theorem )
This gives ∠ BOQ = 2 ∠ OAQ ……………………….(i)
Similarly, ∠ BOP = 2 ∠ OAP ……………………….(ii)
A
P Q
O
B
To prove: ∠ POQ = 2 ∠ PAQ.
9. From (i) and (ii) we get,
∠ BOP + ∠ BOQ = 2 (∠ OAP + ∠ OAQ)
Now,
∠ POQ = 2 ∠ PAQ …………………..(iii)
For the case (iii)where PQ is the major arc
(iii) is replaced by reflex angle
POQ = 2 ∠ PAQ
Proved
A
P Q
O
B
To prove: ∠ POQ = 2 ∠ PAQ.
O
Q
A
P
B
10. Opposite Angles in a Cyclic Quadrilateral
are supplementary
Required to Prove that x + y = 180º
x
2x
y
2y
Draw in radii
The angle at the centre is
TWICE the angle at the
circumference
2x + 2y = 360º
2(x + y) = 360º
x + y = 180º
Opposite Angles in Cyclic Quadrilateral are Supplementary
11. Tangent to a Circle
The tangent at any point of a circle is perpendicular to the
radius through the point of contact
Given: a circle with centre O and a tangent XY to the circle
at a point P.
To prove: OP is perpendicular to XY.
Construction: Take a point Q on XY
other than P and join OQ ,
The point Q must lie outside the circle.
Proof: OQ is longer than the radius
OP of the circle. That is,
OQ > OP
O
X YP Q
12. Tangent to a Circle
Since this happens for every point on the
line XY except the point P, OP is the
shortest of all the distances of the point O to the
points of XY.
So OP is perpendicular to XY
O
X YP Q
13. Tangent to a Circle
The lengths of tangents drawn from an external point to a
circle are equal.
Given: A circle with centre O, a point P
lying outside the circle and two tangents
PQ, PR on the circle from P
To prove: PQ = PR.
Construction: With centre of circle at O,
draw straight lines OA and OB, Draw straight line OP.
O
Q
R
P
14. O
Q
R
P
Tangent to a Circle
Proof: In right triangles OQP and ORP,
OQ = OR .......................(radii of the same circle)
∠ OQP = ∠ ORP ………………..(right angles)
OP = OP ………………(Common)
Therefore,
Δ OQP ≅ Δ ORP ……..(RHS)
This gives
PQ = PR …………..(CPCT)
15. The End
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