**Circle:** The closed curved line at affixed distance from a fixed point is called circle. The fixed point

is the centre of the circle and the fixed distance is called radius of the circle.

### In the figure O is the centre and OA is radius of the circle.

**Diameter**: A chord of a circle which passes

through the centre is called diameter. Any

chord that passes through the centre is a

diameter. The diameter is the largest chord

of the circle. Half of the diameter is the radius

of the circle. Obviously, diameter is twice the radius.

**Circumference**: The complete length of the circle is called its circumference.

In the figure AB is the diameter and AOB is the circumference.

**Exercise**: **prove that the perpendicular from the centre of circle to a chord bisects the chord.**

General enunciation: To prove that the perpendicular from the centre of a circle to a chord bisects the chord.

**Particular enunciation**: let AB be chord other than diameter of a circle with centre O and we draw a perpendicular OD from the centre O to this chord. We have to prove that OD bisects the chord AB at D. i.e. AD=BD

Construction: we join O,A and O,B.

**Proof**: Since OD⊥AB,

We have ∠ODA=∠ODB=1 right angle

So both of ∆ODA and ∆ODB are right angled triangle

In the right angled triangle ∆ODA and ∆ODB

Hypotenuse OA= hypotenuse OB [both are radius of the same circle ]

and OD=OD [common side]

so ∆ODA≌∆ODB [HS theorem]

therefore AD=BD [proved]

**Exercise**: **prove that, if two chords of a circle bisect each other , their point intersection will be the centre of the circle.**

General enunciation: We have to prove that, if two chords of a circle bisect each other, their point of intersection will be the centre of the circle.

**Particular enunciation**: Let in the circle ABCD the chords AB and CD bisect each other at E.

We have to prove that the point E is the centre of the circle ABCD.

**Construction**: Let the centre of the circle is O rather than E. We join O,E.

**Proof**: O be the centre of the circle. E be the midpoint of the chords AB and CD respectively.

∴OE⊥AB and OE⊥CD

The line segment joining the centre of a circle to midpoint of a chord other than diameter, is perpendicular to the chord.

So ∠OEC = 1 right angle

AB and CD are two intersector straight lines.

Both ∠OEA and ∠OEC are not equal to 1 right angle.

∴ No point other than E can be the centre of the circle.

∴E be the centre of the circle. [proved]

**Exercise**: **Prove that the line joining the midpoints of two parallel chords passes through the centre and is perpendicular to the two chords.**

General enunciation: we have to prove that the line joining the midpoints of two parallel chords passes through the centre and is perpendicular to the two chords.

**Particular enunciation**: Let in the circle of centre O, AB and CD are two parallel chords. E and F are the midpoints of AB and CD respectively. The line segment EF is the joining line of the midpoints E and F. It is required to prove that EF passes through the centre and is perpendicular to the chords.

Construction: We join O and F.

Proof: O be the centre of the circle and E and F be the midpoints of the chords AB and CD.

∴OE⊥AB and OF⊥CD

The line segment joining the centre of the circle to midpoint of a chord other than diameter, is perpendicular to the chord.

So ∠OEB = 1 right angle

And ∠OFD= 1 right angle

∴∠OEB=∠OFD.

The two angles ∠OEB and ∠OFD are corresponding and equal.

∴OF and OE will be the transversals [AB∥CD]

i.e. lie on the same straight line [O be the centre of the circle]

Therefore EF passes through the centre (proved)