** A circle is a labeled by its center point: O means the circle with center point O**

**Diameter**

A line segment, generally denoted by the variable, that connects two points on the circles and passes through the center of the circle

**Radius:- **A line segment, generally denoted by the variable r from the center of the circle to any point on the circle.

**Chord: **A line Segment joining two points on the circle segment C is a chord.

**Tangent:- **A line that touches only one point on the circumference of the circle. A line drawn a tangent to a circle is perpendicular to the radius at the point of contact.

**Circumference:- **The distance around a circle is called the circumference.

An Arc is a portion of the circumference of a circle. The shorter distance between A an B along the circle is called the minor arc.

The longer distance A and B is the major arc.

**Arc length:- **For an arc with a central angle measuring θ Degree:

**Length = ( θ / 360 ) (Circumference)**

**= (θ / 360 ) ( π d )**

**Example: What is the length of arc Abc of the circle with center O shown?**

**Ans:- **Arc length= (Q / 360 ) ( π d )= (60 / 360 ) (π (2))

12 π / 6 = 2 π

**Area of circles: **The area of circles is given by the formula A= π r^{2}

A sector is a portion of the circle that is bounded by two radii and an arc.

In a sector Whose central angle measures by Q degrees.

Area of sector= (Q / 360) x (Area of circle)

**Example:- What is the area of sector AOC in the circle with the center O shown?**

**Ans) **Area of sector AOC= ( 60 / 360 ) (36π ) = ( 6 π )

**PROPERTIES:–**

- The perpendicular from the center of a circle to a chord bisects the chord and vice versa.

**If ∠OCB=90 °, THEN AC=BC**

**If AC=BC , then ∠OCB= 90°**

**2)** Equal chords of a circle are equidistant from the center. Conversely, chords equidistant from the center are always equal.

**3)** Any two angles in the same segment are equal. Thus **∠ACB = ∠ADB**

**4)**The angle subtended by an arc at the center of the circle is twice the angle subtended by the same arc at any other point on the circle.

**∠AOB= 2∠ACB**

**5)**The angle subtended by a semicircle is a right angle. Conversely, the arc of a circle subtending a right angle at any point on the circle is a semi-circle.

**If AB is a diameter, then ∠ACB = 90°**

**IF ∠ACB = 90° , then AB is a diameter.**

**6)**Tangent drawn from common external point to a circle are equal.

**Q):- A 5 by 12 rectangle is inscribed in a circle. What is the circumference of the circle?**

**Ans:-**

∠BCD= 90°

BD is a diameter (Because The arc of a circle subtending a right angle at any point on the circle is a semi-circle)

In BCD , BC^{2} + DC^{2} = BD^{2} (by Pythagoras theorem)

BD = 13

Circumference of the circle = 2π (Radius) = 13 π

**Q) In the figure shown below, if the radius of a circle with centre P is three times the radius of circle with centre A, ∠BAC=∠QPR, and the shaded area of the circle with centre A is 3 π SQUARE units, then what is the area of the shaded part of circle with centre P?**

**Ans) **Let ∠BAC =∠QPR = Q°

The shaded area of circle A = 3 π sq. units = ( Q / 360° ) π (AC)^{2}

PQ = 3AC

= ( Q / 360° ) π (PQ)^{2}

= ( Q / 360° ) π (3AC)^{2}

= (3π x 9)

= 27π sq. units

**Ques 28: -Two congruent, adjacent circles are cut out of a 16 by 8 rectangle. The circles have the maximum diameter possible. What is the area of the paper remaining after the circles have been cut out?**

**Ans) **For the circles, the diameter of the circle is the same as the width of the rectangle.

Remaining area = Area of rectangle – 2 x area of circle

Area of rectangle = Length x Breadth = 8 x 16 = 128

radius of circle = 4

Area of circle = π (4)^{2 }= 16 π

Remaining area = (128- 2x(16π) ) = ( 128 – 32 π) sq. units

**Ques: The Figure shows an equilateral triangle, where each vertex is the center of a circle. Each circle has a radius of 20. What is the area of the shaded region?**

**Ans:- **Side of equilateral triangle =40

Area of equilateral triangle= √3/4 (40)^{2}

=√3/4 (1600) = (400 √3)

Area of three 60° sectors= 3 x (60/360) (π) (r^{2})

=(1/2) π (400) = 200π

Area of shaded Region= Area of equilateral triangle- area of three 60° degree sectors

= (400√3 – 200 π ) sq. units