A circle is the collection of all the points which are at a fixed distance from a fixed

point in a plane.

The fixed point is called centre of the circle.

The fixed distance is called the radius of the circle.

It is the part of plane which is within the circle.

- The distance between any point in interior region and centre is less than

the radius of the circle.

The part of plane which is out side the circle.

- The distance between any point in exterior region and centre is

greater than the radius of the circle.

The line segment joining any two points of the circle is called chord.

The diameter is chord passing through the centre.

It is the longest chord of the circle.

Diameter = 2 X r

A piece of circle between any two points is called arc.

The angle formed at centre by joining end points of Chord /arc and centre is called centre angle of arc/chord.

If central angle of an arc is acute ,the arc is called minor arc.

If central angle of an arc is obtuse, the arc is called major arc.

The diameter divides part of circles in two equal arcs, called semicircles.

$Forsemicircleangleformedbyarcsatcentreis{180}^{\circ}.$The interior part of circle, the region between chord and arc is called segment.

The region between major arc and chord is called major segment.

The region between minor arc and chord is called minor segment.

The region between two radii and an arc is called sector.

The region between two radii and an arc is called sector.

The region between minor arc and radii is called minor sector.

Equal chords of a circle subtend equal angles at the centres.

O is centre of the circle.

Chord AB = chord CD

AB = CD …..(Given)

AO = CO …………….( Radii of same circle)

BO = DO ……………( Radii of same circle)

$\u25b3AOB\cong \u25b3COD................(SSSrule)$ $So,\angle AOB=\angle COD.................\left(CPCT\right)$If angles subtended by the chord of a circle at the centre are equal, then the Chords are equal.

O is centre of the circle.

$\angle AOB=\angle COD$

AO = CO …………….( Radii of same circle)

BO = DO ……………( Radii of same circle)

$\u25b3AOB\cong \u25b3COD................(SOSrule)$ $\therefore AB=CD.................\left(CPCT\right)$Perpendicular from centre to a chord bisects the chord.

**Given**:

O is centre of the circle.

AB is chord.

$OM\perp AB$OM bisects AB i.e. AM = BM

AO = BO ……….(Radii)

OM = OM ………..(Common)

$\angle AMO=\angle BMO$ $\u2206AOM=\u2206BOM\dots \dots .(R\u2013HSrule)$So, AM = BM …………….. (CSCT)

The line joining centre and midpoint of chord is perpendicular to chord.

O is centre of the circle.

AB is chord.

M is midpoint Of AB,

$\therefore AM=BM$AO = BO ………(Radii)

OM = OM……. (Common_

AM = BM………(Given)

$\u2206AOM=\u2206BOM\dots \dots \dots \dots (SSSrule)$ $So\angle AMO=\angle BMO\left(CPCT\right)$But

$\angle AMO+\angle BMO={180}^{\circ}(Linearpair)$ $2\angle AMO={180}^{\circ}\angle AMO={90}^{0}$$SoOM\perp AB$

There is one and only one circle passing through three non-collinear points.

**Remark**: There is a unique circle passing through vertices of the triangle.

Equal chords are equidistant from centre.

AB and CD are two chords.

AB = CD

$OM\perp AB,ON\perp CD$ON = OM

$\therefore AN=\frac{1}{2}AB$

$CM=\frac{1}{2}CD$

But, AB = CD

$\therefore AN=CM...\left(1\right)$

$Inright\u2206OCMand\u2206OAN$

AN = CM …(From 1)

OA = OC …(Radii)

$\u2206OCM\cong \u2206OAN.....(R.H.Srule)$

$\therefore ON=OM.....\left(CPCT\right)$

Chords equidistant from centre are equal.

AB and CD are two chords.

$OM\perp AB,ON\perp CD$ON = OM (As chords are equidiastantfrom the centre)

AB = CD

**Proof:**

ON = OM …(Given)

OA = OC …(Radii)

$\u2206OCM\cong \u2206OAN....\dots ..(RHSrule)$

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