It is the set of all points in a plane that are at a given distance from a given point, the center; equivalently it is the curve traced out by a point that moves so that its distance from a given point is constant. The distance between any of the points and the center is called the radius.

Perimeter of the circle is called its circumference.

*Circumference of a circle*

*Area of a circle*

In any circle, there are three types of points (refer to the diagram):

*Exterior Points:*These are points outside the circle. The distance of these points from the center is greater than the length of the radius of the circle.*Interior Points:*These are points inside the circle.The distance of these points from the center is less than the length of the radius of the circle.*Points on the circumference:*These are the points on the circumference. In this case, the distance of the point from the center of the circle is equal to the length of the radius of the circle.

*Concentric circles:* Two or more circles are called concentric if they have the same center but different radii.

*Equal circles:* The circles are said to be equal or congruent if they have the same radii.

*Circumscribed circle: *A circle that passes through all the vertices of a polygon is called a circumscribed circle. The center of this circle is called circumcenter and the polygon is called inscribed polygon.

*Inscribed circle:* A circle that touches all the sides of the polygon is called inscribed circle of the polygon. The center of this circle is called incenter and the polygon is called the circumscribed polygon.

**Theorem 4: ***A straight line drawn from the center of the circle to bisect a chord, which is not the diameter, is perpendicular to the chord.*

Given: is the center of the circle

To Prove:

Proof: Consider

(radius of the same circle)

is common

(given)

Therefore (S.S.S postulate)

Therefore (corresponding angles in congruent triangles are equal)

Hence

**Hence Proved.**

**Theorem 5:** *A perpendicular to a chord, from the center of the circle, bisects the chord.*

Given: is the center of the circle.

To Prove:

Proof: Consider

(given)

is common

(radius of the circle)

Therefore (R.H.S postulate)

Therefore (congruent parts of the triangles are equal)

Therefore bisects and

**Hence Proved. **

*Note: Greater the size of the chord, closer it is to the center of the circle and smaller the size of the chord, farther away it is from the center of the circle. In the adjoining figure, *

**Theorem 6: ***There is only one circle, and only one, which passes through three given points which are not in a straight line.*

Given: Three points which are not in a straight line.

To Prove: Only one circle can be drawn through .

Construction: Join . Draw perpendicular bisector of . The perpendicular bisectors meet at .

Proof: lies on the perpendicular bisector of

Also since, likes on the perpendicular bisector of

Therefore

Which means is equidistant from .

So if you were to draw a circle with as the center and as the radius, the circle will pass through .

The perpendicular bisector of intersect each other only at point .

is the only point which is equidistant from .

Hence, one and only one circle can be drawn through which are not in a straight line.

**Hence Proved.**

*Note:*

*The perpendicular bisector of any chord of a circle will always pass through the center of this circle.**Perpendicular bisector of any two chords of a circle always intersect at the center of the circle.*

**Theorem 7: ***Equal chords of a circle are equidistant from the center.*

Given: Chord Chord

To Prove: Chords and Chords are equidistant from the center of the circle. i.e. Prove

Proof: Join

( Perpendicular from the center will always bisect the chord)

Similarly,

But (given)

Therefore

Now consider

(proved above)

(radius of the circle)

Therefore (R.H.S postulate)

**Hence Proved.**

**Theorem 8 ***(Converse of Theorem 7)*** :** *Chords of circle which are equidistant from the center of the circle are equal in length*.

Given: Chords are equidistant. This means that and

To Prove:

Proof: Consider

(Given)

(radius of the same circle)

()

Therefore (R.H.S postulate)

Therefore (Corresponding sides of congruent triangles are equal).

We know that perpendicular from the center of the circle will bisect the chord.

Therefore

.

**Hence proved. **

*Note: If parallel chords are on the same side of the circle, then the line through the midpoints of the chord will pass through the center.*

*Note: If two equal chords intersect inside the triangle at point , then we have i) and ii) i.e the corresponding segments of the chords are equal.*