You might have seen many objects that are circular in shape. A circle is defined as the collection of all the points on a plane that are at equal distance from a given fixed point on the plane. This fixed point is called centre of the circle and the fixed distance is called the radius.
A line segment joining the centre of a circle with any point on its circumference is called the radius of the circle.
A line that joins two points on the circumference of a circle is called a chord. A chord that passes through the centre of a circle is called the diameter of the circle. A diameter divides a circle into two equal parts, each is called a semicircle. Diameter is the longest chord of a circle. The diameter of a circle is twice the radius.
The part of the circumference of a circle between two given points is called an arc.
A chord divides a circular area into two parts called segments they are major segment and minor segments. The region between two radii of a circle and any of the arcs between them is called a sector. The diameter of a circle divides it into two equal segments.
You know that the perpendicular from a point to a line segment is the shortest distance between them. A line that joins two points on the circumference of a circle is called a chord. A chord passing through the centre of a circle is called the diameter. The longest chord of a circle is the diameter.
Theorem: The perpendicular from the centre of a circle to a chord bisects the chord.
Given: OB ⊥ AC.
To prove: AB = BC.
Construction: Join OA and OC.
Proof: In triangles OBA and OBC,
∠OBA = ∠OBC = 90o (Since OB ⊥ AC.)
OA = OC (Radii of same circle)
OB = OB (Common side)
ΔOBA 
ΔOBC (By RHS congruence rule)
AB = BC
Thus, OB bisects chord AC.
Hence, the theorem is proved.
Theorem: The line drawn from the centre of a circle to bisect a chord is perpendicular to the chord.
Given: AB = BC
To prove: OB ⊥ AC.
Construction: Join OA and OC.
Proof: In triangles OBA and OBC,
AB = BC (Given)
OA = OC (Radii of same circle)
OB = OB (Common side)
Δ OBA Δ OBC (SSS congruence rule)
∠OBA = ∠OBC
∠OBA + ∠OBC = ∠ABC = 180o
Since ∠OBA = ∠OBC,
2 x ∠OBC = 180o
∠OBC = 
= 90o
∠OBC = 90o = ∠OBA
∴OB ⊥ AC
Theorem: Equal chords of a circle are equidistant from the centre of the circle.
Theorem: Chords equidistant from the centre of a circle are equal in length.
Let AB be any chord of a circle with the centre O. Then is called the angle subtended by the chord at the centre of the circle. As the chord moves away from the centre, its length and the angle subtended by it at the centre decreases. On the other hand, as it moves closer to the centre, its length and the angle subtended by it at the centre increases.
Theorem: Chords that subtend equal angles at the centre of a circle are equal in length.
Theorem: Equal chords of a circle subtend equal angles at the centre.
Given: AB = PQ
To prove: ∠AOB = ∠POQ
Proof: In triangles AOB and POQ,
AB = PQ (Given)
OA = OP (Radii of same circle)
OB = OQ (Radii of same circle)
Δ AOB Δ POQ (SSS congruence rule )
∠AOB= ∠POQ
Hence, the theorem is proved.
Converse of the theorem:
Theorem: Chords that subtend equal angles at the centre of a circle are equal in length.
Given: ∠AOB= ∠POQ
To prove: AB = PQ
Proof: In triangles AOB and POQ,
∠AOB= ∠POQ (Given)
OA = OP (Radii of same circle)
OB = OQ (Radii of same circle)
Δ AOB Δ POQ (SAS congruence rule)
AB = PQ
Hence, the theorem is proved.
Every day, you come across many things circular in shape. The collection of all that points in a plane that 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.
A part of a circle is called an arc. Arcs of a circle that superimpose each other completely are called congruent arcs. A segment with its endpoints on a circle is called a chord. A diameter is the longest chord. If two arcs of a circle are congruent, then their corresponding chords are equal. Conversely, if two chords of a circle are equal, then their corresponding arcs are congruent.
Corresponding Arcs Of Two Equal Chords Of A Circle Are Congruent
Theorem: Congruent arcs of a circle subtend equal angles at the centre.
Given: Two congruent arcs AB and CD.
To prove: ∠ AOB = ∠ COD
Construction: Draw chords AB and CD.
Proof: The angle subtended by an arc at the centre is equal to the angle subtended by its corresponding chord at the centre.
In the given figure,
AB = CD (Chords corresponding to congruent arcs of a circle are equal)
∠ AOB = ∠ COD (Equal chords subtend equal angles at the centre)
Hence, the theorem is proved.
Theorem: The angle subtended by an arc at the centre is double the angle subtended by the arc at any point on the circle.
Given:
Arc AB.
Point C on the circle is outside AB.
To prove: ∠ AOB = 2 x ∠ ACB
Construction: Draw a line CO extended till point D.
Proof: In Δ OAC in each of these figures,
∠ AOD = ∠ OAC + ∠ OCA (Exterior angle of a triangle is equal to sum of two opposite interior angles)
OA = OC (Radii of same circle)
Thus, ∠ OAC = ∠ OCA (Angles opposite equal sides of a triangle are equal)
∠ AOD = ∠ OAC + ∠ OCA
⇒∠ AOD = 2 x ∠ OCA
Similarly, in Δ OBC,
∠ BOD = 2 x ∠ OCB
∠ AOD = 2 x ∠ OCA
and ∠ BOD = 2 x ∠ OCB
⇒∠ AOD + ∠ BOD = 2∠ OCA + 2∠ OCB
∠ AOD + ∠ BOD = 2 x (∠ OCA + ∠ OCB)
Or ∠ AOB = 2 x ∠ ACB
Hence, the theorem is proved.
Theorem: Angles subtended by an arc at all points within the same segment of the circle are equal.
Given:
An arc AB.
Points C and D are on the circle in the same segment.
To prove: ∠ ACB = ∠ ADB
Proof: By the theorem that the angle subtended by an arc at the centre is double the angle subtended by the arc at any point on the circle:
∠ AOB = 2 x ∠ ACB
Also, ∠ AOB = 2 x ∠ ADB
∴∠ ACB = ∠ ADB
Hence, the theorem is proved.
All angles formed in a semi circle are right angles
You can draw a circle passing through three non-collinear distinct points. The points that lie on a circle are called concyclic points. So we can say three non-collinear points are always concyclic.
Theorem: If a line segment joining two points subtends equal angles at two other points on the same side of the line segment then all the four points are concyclic.
Given: Line segment AB.
Mark two points C and D such that ACB = ∠ADB.
To prove: A, B, C and D are concyclic points.
Draw a circle through points A, B and C.
Assume that the circle drawn through points A, B and C does not pass through D, and intersects AD at D’.
Proof: If A, B, C and D’ are concyclic:
∠ACB = ∠AD’B (Angles subtended by a chord in the same segment of a circle)
∠ACB = ∠ADB (Given)
∴ ∠AD’B = ∠ADB
Or D’ coincides with D.
Thus, A, B, C and D are concyclic points.
Hence, the theorem is proved.
A quadrilateral whose vertices lie on a circle is called a cyclic quadrilateral.
In a cyclic quadrilateral, the sum of the opposite angles is always equal to 180o.
If the sum of the opposite angles of a quadrilateral is 180o, then the quadrilateral is cyclic.
AREA OF CIRCLE= ∏r^2
Parimeter=2∏r
Note Blast 