Before you start learning about circles related to 9th grade curriculum, please revise what we learnt in 8th grade. 8th Grade Revision – Circles.

Congruence of Circle and Arcs

Congruent Circles:

Two circles are said to be congruent if and only if either if them can be super imposed on the other so as to cover it exactly.

This means that two circles are congruent if and only if their radii are equal. i.e. circles and are congruent if .

Congruent Arcs:

Two arcs of a circle are congruent if either of them can be super imposed on the other so as to cover it exactly.

This happens if degree measures of two arcs are the same. Therefore we can say that two arcs of a circle (or of congruent circles) are said to be congruent if and only if they have the same degree measure.

If two arcs and are convergent arcs if a circle , we write .

Thus,

Important results and their proofs on congruent arcs and chords:

**Theorem 1:** **If two arcs of a circle (or of congruent circles) are congruent, then the corresponding chords are equal.**

**Explanation:** Follow the diagram shown:

Given: Two circles and . They are congruent hence their radius is equal.

Also .

Construction: Join and to in the first circle and and to in the second circle.

To Prove: (i.e. corresponding chords are equal).

There can be two cases: When and are minor arcs (case 1) and When and are major arcs (case 2).

Case 1: and are minor arcs

Consider and

(radius of the two circles are equal)

Also

(Since )

Hence (by SAS theorem)

. Hence proved.

Case 2: and are major arcs

From *case 1*, we proved that that

Hence

**Theorem 2: ****If two chords of a circle (or of congruent circles) are equal, then the corresponding arcs (major, minor or semi-circular) are congruent.**

**Explanation: **Follow the diagram shown:

Given: Chords of circle and of circle are equal.

Construction: Join and to in the first circle and and to in the second circle. and are not diameters.

To Prove: where they can be both minor, major or semi-circular arcs.

Proof: There are three possible cases.

Case 1: When and are diameters.

In this case, and are semi circles of equal radii. Hence they are congruent.

Case 2: When and are minor arcs.

Consider and

Given:

Therefore (By S.S.S criterion)

Case 3: When and are major arcs

and are minor arcs

Therefore in all three cases

**Theorem 3: The perpendicular from the center of a circle to a chord bisects the chord.**

Given: is a chord in Circle and

To Prove:

Construction: Join and

Proof: Consider and

(given)

is common

Therefore (By S.S.A criterion)

Therefore

Hence bisects

**Theorem 4 (converse of Theorem 3): If the line joining the center of the circle to the mid point of the chord is perpendicular to the chord.**

Given: is the midpoint of

is a chord in circle

To Prove:

Proof: Consider and

is common

(given)

Therefore (By S.S.S criterion)

Therefore

Since (linear pairs)

**Theorem 4.1: The perpendicular bisectors of two chords of a circle intersect at its center.**

Given: and are chords in circle

( is the midpoint of )

( is midpoint of )

and

Let and meet at

To Prove: coincides with

Construction: Join and

Proof: Since is the mid point of

(Theorem 4)

is perpendicular bisector of

and are perpendicular bisector of

lies on

Similarly, is the mid point of

is perpendicular bisector of (Theorem 4)

and are perpendicular bisector of

lies on

Therefore, we can say (point of intersection of and ) coincides with (center of the circle)

Therefore, Perpendicular bisector of and will intersect at the center of the circle.

**Theorem 5: There is one and only one circle passing through three non-collinear points.**

Given: and are three non-collinear points.

To Prove: One and only one circle passes through and

Construction: Join and

Draw perpendicular bisector of and also draw perpendicular bisector of

Let and intersect at

Proof: Since lies on perpendicular bisector of

Similarly,

(radius of the circle)

Now taking as the center and as the radius, draw a circle.

You will see that the circle passes through the points and .

We now need to prove that this is the only circle passing through the points and .

Now let us assume that there is another circle with center latex P, Q $ and .

Then will lie on the perpendicular bisector of of and of

Since two lines cannot intersect in two points, must coincides with

Hence only one and one circle can pass through three non-collinear points.

*Note 1: In a case where the three points are collinear, then a single circle cannot pass through all of them.*

*Note 2: This theorem also implies that there will be a unique circle that passes through the vertices of a triangle. This circle is called circum-circle of the triangle and the center of the circle is called circum-center.*

*Note 3: Infinite number of circles can be drawn through one point.*

*Note 4: Infinite number of circles can be drawn through two points.*

**Theorem 6: If two chords AB and AC of a circle with center O are equal then the center of the circle lies on the angle bisector of .**

Given: are chords of circle

To Prove: lies on the angle bisector of

Construction: Join . Draw bisector of

Proof: Consider and

(given)

is common

(By construction)

Therefore (By S.A.S criterion)

and

Since

is perpendicular to and it bisects

passes through the center (By theorem 4)

**Theorem 7 (converse of Theorem 6): If two chords and of a circle with center are such that the center lies on the bisector of , then i.e. chords are equal.**

Given: and are chords of circle

To Prove:

Construction: Join and

Proof: In

(radius)

(angles opposite to equal sides are equal)

Similarly, in

(radius)

(angles opposite to equal sides are equal)

But

Consider and

is common

Therefore

(since corresponding parts of congruent triangles are equal)

**Theorem 8: If two circles intersect in two points, then the line through the centers of the circles is perpendicular bisector of the common chord.**

Given: Two circles and

They intersect at and

Construction: Join and . Also join and

To Prove: is perpendicular bisector of

Proof: Consider and

is common

Therefore (By S.S.S criterion)

Let and intersect at

Consider and

is common

Therefore (By S.A.S criterion)

and

We know

Thus and

Hence is perpendicular bisector of