What Is a circle….. A circle is a collection of all the points in a plane , which are at fixed distance from a fixed point in the plane. The fixed point is called the center of the circle and the fixed distance is called the radius of the circle

Radius, Diameter and Circumference The Radius is the distance from the center to the edge. The Diameter starts at one side of the circle, goes through the center and ends on the other side. The Circumference is the distance around the edge of the circle.

If that line passes through the center it is called a Diameter. Lines A line that goes from one point to another on the circle's circumference is called a Chord. If that line passes through the center it is called a Diameter. A line that "just touches" the circle as it passes by is called a Tangent. And a part of the circumference is called an Arc.

Slices There are two main "slices" of a circle. The "pizza" slice is called a Sector. And the slice made by a chord is called a Segment.

Common Sectors The Quadrant and Semicircle are two special types of Sector: Quarter of a circle is called a Quadrant. Half a circle is called a Semicircle.

Inside and Outside A circle has an inside and an outside (of course!). But it also has an "on", because we could be right on the circle. Example: "A" is outside the circle, "B" is inside the circle and "C" is on the circle.

Proof for theorm 10.1   Let two equal chords PQ and RS of a circle with centre O are of equal length. It is required to prove that ∠POQ = ∠ROS. In triangles POQ and ROS, OP = OR   (Radii of a circle) OQ = OS   (Radii of a circle) PQ = RS   (Given) Therefore, ΔPOQ ≅ ΔROS (SSS rule) ⇒ ∠POQ = ∠ROS   (Corresponding parts of congruent triangles)

Proof for theorm 10.2 angle Subtended By An Arc Of A Circle Given an arc PQ of a circle subtending angles POQ at the O and PAQ at a point a on the remaining part of the circle . We need to prove that angle POQ = 2 angle PAQ .

CASE- (I) A O B Q P

Case -(II) A P o B Q

Case – (III) Q P O B

Consider the three different cases as given in fig 1. 3 Consider the three different cases as given in fig 1.3.In (I) , arc PQ is minor , in (II) , arc PQ is a semicircle and in (III) , arc PQ is major.   Let us begin by joining AO and extending it to a point B.

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 triangle OAQ ,   OA=OQ (Radii of a circle) Therefore , ^OAQ=^OQA (Theorem) This gives ^BOQ=2^OAQ (I) Similarly , ^BOP=2^OAP (II)

From (I) and (II) , ^BOP +^BOQ = 2(^OAP+ ^OAQ)   This is same as ^POQ = 2 ^PAQ (III) For the case (III) , where PQ is the major arc , (III) is replaced by reflex angle POQ = 2 ^PAQ.