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Home / Geometry / CIRCLE| Part-1

# CIRCLE| Part-1

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]

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)

## If two triangles have the three sides of the one equal to the three sides of the other, each to each, then they are equal in all respects.

If two triangles have the three sides of the one equal to the three sides ...