CIRCLES

Introduction to Circles

Circles

- The set of all the points in a plane that is at a fixed distance from a fixed point makes a circle.
- A Fixed point from which the set of points are at fixed distance is called centre of the circle.
- A circle divides the plane into 3 parts: interior (inside the circle), the circle itself and exterior (outside the circle)

Radius

- The distance between the center of the circle and any point on its edge is called the radius.

Tangent and Secant

A line that touches the circle at exactly one point is called its tangent. 
A line that cuts a circle at two points is called as a secant.


In the above figure: PQ is the tangent and AB is the secant.

Chord

-The line segment within the circle joining any 2 points on the circle is called the chord.

Diameter

- A Chord passing through the center of the circle is called the diameter.
- The Diameter is 2 times the radius and it is the longest chord.

Arc

- The portion of a circle(curve) between 2 points is called an arc.
- Among the two pieces made by an arc, the longer one is called major arc and the shorter one is called minor arc.
 

Circumference

The perimeter of a circle is the distance covered by going around its boundary once. The perimeter of a circle has a special name: Circumference, which is $\pi$ times the diameter which is given by the formula $2\pi r$

Segment and Sector

- A circular segment is a region of a circle which is "cut off" from the rest of the circle by a secant or a chord.
- Smaller region cut off by a chord is called minor segment and the bigger region is called major segment.

- A sector is the portion of a circle enclosed by two radii and an arc, where the smaller area is known as the minor sector and the larger being the major sector.
- For 2 equal arcs or for semicircles - both the segment and sector is called the semicircular region.

Circles and Their Chords

Theorem of equal chords subtending angles at the center.

- Equal chords subtend equal angles at the center.

Proof : AB and CD are the 2 equal chords. 
In $\Delta AOBand\Delta COD$
$OB=OC$ [Radii]
$OA=OD$ [Radii]
$AB=CD$ [Given] 
$\Delta AOB\cong \Delta COD$ (SSS rule)
Hence, $\angle AOB=\angle COD$ [CPCT]

Theorem of equal angles subtended by different chords.

- If the angles subtended by the chords of a circle at the center are equal, then the chords are equal.

 
Proof : In $\Delta AOBand\Delta COD$
$OB=OC$ [Radii]
$\angle AOB=\angle COD$ [Given] 
$OA=OD$ [Radii]
$\Delta AOB\cong \Delta COD$(SAS rule)
Hence, $AB=CD$ [CPCT]

Perpendicular from the center to a chord bisects the chord.

Perpendicular from the center of a circle to a chord bisects the chord.

Proof: AB is a chord and OM is the perpendicular drawn from the center.
From $\Delta OMBand\Delta OMA,$
$\angle OMA=\angle OMB={90}^0$
$OA=OB$ (radii)
$OM=OM$ (common)
Hence,$\Delta OMB\cong \Delta OMA$ (RHS rule)
Therefore $AM=MB$ [CPCT]

A Line through the center that bisects the chord is perpendicular to the chord.

- A line drawn through the center of a circle to bisect a chord, is perpendicular to the chord.

Proof: OM drawn from center to bisect chord AB .
From $\Delta OMAand\Delta OMB,$
$OA=OB$ (Radii)
$OM=OM$ (common)
$AM=BM$ (Given)
Therefore, $\Delta OMA\cong \Delta OMB$ (SSS rule)
$\Rightarrow \angle OMA=\angle OMB$ (C.P.C.T)
But, $\angle OMA+\angle OMB={180}^0$
Hence, $\angle OMA=\angle OMB={90}^0$
$\Rightarrow OM\perp AB$

Circle through 3 points

- There is one and only one circle passing through three given noncollinear points.
- A unique circle passes through 3 vertices of a triangle ABC called as the circumcircle. The centre and radius are called the circumcenter and circumradius of this triangle, respectively.

Equal chords are at equal distances from the center.

Equal chords of a circle(or of congruent circles) are equidistant from the centre (or centres).


Proof : Given, $AB=CD,O$ is the centre. Join $OA$ and $OC$.
Draw, $OP\perp AB,OQ\perp CD$
In $\Delta OAPand\Delta OCQ$,
$OA=OC$ (Radii)
$AP=CQ$ ( AB = CD $\Rightarrow \frac{1}{2}AB=\frac{1}{2}CD$, since OP and OQ bisects the chords AB and CD.)
$\Delta OAP\cong \Delta OCQ$ (RHS rule)
Hence, $OP=OQ$ (C.P.C.T.C)

Chords equidistant from center are equal

Chords equidistant from the center of a circle are equal in length.


Proof : Given OX = OY (The chords AB and CD are at equidistant)
$OX\perp AB,OY\perp CD$
In $\Delta AOXand\Delta DOY$
$\angle OXA=\angle OYD$  (Both ${90}^0$)
$OA=OD$ (Radii)
$OX=OY$ (Given)
$\Delta AOX\cong \Delta DOY$ (RHS rule)
Therefore $AX=DY$ (CPCT)
Similarly $XB=YC$ 
So, $AB=CD$

Circles and Quadrilaterals

Angle subtended by an arc of a circle on the circle and at the center

The angle subtended by an arc at the centre is double the angle subtended by it on any part of the circle.

Here PQ is the arc of a circle with centre O, that subtends $\angle POQ$ at the centre. 
Join AO and extend it to B.
In $\Delta OAQ$
$OA=OQ$..... [Radii]
Hence, $\angle OAQ=\angle OQA$....[Property of isosceles triangle]
Implies $\angle BOQ=2\angle OAQ$ .....[Exterior angle of triangle = Sum of 2 interior angles]
Similarly, $\angle BOP=2\angle OAP$
$\Rightarrow \angle BOQ+\angle BOP=2\angle OAQ+2\angle OAP$
$\Rightarrow \angle POQ=2\angle PAQ$  
Hence proved

Angles in same segment of a circle.

-Angles in the same segment of a circle are equal.



Consider a circle with centre O.
$\angle PAQ$ and $\angle PCQ$ are the angles formed in the major segment PACQ with respect to the arc PQ.
Join OP and OQ
$\angle POQ=2\angle PAQ=2\angle PCQ$ .....[ Angle subtended by an arc at the centre is double the angle subtended by it in any part of the circle]
$\Rightarrow \angle PCQ=\angle PAQ$
Hence proved

Angle subtended by diameter on the circle

- Angle subtended by diameter on a circle is a right angle.(Angle in a semicircle is a right angle)



Consider a circle with center O, POQ is the diameter of the circle.
$\angle PAQ$ is the angle subtended by diameter PQ at the circuference.
$\angle POQ$ is the angle subtended by diameter PQ at the center.
$\angle PAQ=\frac{1}{2}\angle POQ$... [Angle subtended by arc at the centre is double the angle at any other part] 
$\angle PAQ=\frac{1}{2}\times {180}^0={90}^0$
Hence proved

Line segment that subtends equal angles at two other points

- If a line segment joining two points subtends equal angles at two other points lying on the same side of the line containing the line segment, the four points lie on a circle.(i.e they are concyclic)

Here $\angle ACB=\angle ADB$ and all 4 points A,B,C,D are concyclic.

Cyclic Quadrilateral

- A Quadrilateral is called a cyclic quadrilateral if all the four vertices lie on a circle.



In a circle, if all four points A, B, C and D lie on the circle, then quadrilateral ABCD is a cyclic quadrilateral.

Sum of opposite angles of a cyclic quadrilateral

- If sum of a pair of opposite angles of a quadrilateral is 180 degree, the quadrilateral is cyclic.

Sum of pair of opposite angles in quadrilateral

- The sum of either pair of opposite angles of a cyclic quadrilateral is 180 degree