# Circle (Geometry)

See how to solve circle problems

(arc/segments/angles/area).

25 examples and their solutions.

## Circumference of a Circle

### Formula

C = 2πr

C = πd

### Example

Circumference?

Solution C = 2π⋅5

= 10π

Close

### Example

Circumference?

Solution C = π⋅13

= 13π

Close

## Measure of an Arc

### Definition

mAB = θ

### Example

mAB = ?

Solution mAB = 60

Close

## Length of an Arc

### Formula

### Example

Length of AB = ?

Solution l = 2π⋅6⋅120360

= 12π⋅13

= 4π

Close

## Chord of a Circle

### Definition

whose endpoints are on the circle.

### Property

perpendicularly bisects the chord.

### Example

## Inscribed Angle

### Definition

### Formula

θ = 12mAB

mAB: intercepted arc

### Example

↓

x = 12⋅80

= 40

Close

### Example

mABC = ?

Solution 12⋅mABC = 100

mABC = 200

Close

### Example

7x + 1 = 3x + 29 - [1]

4x = 28

x = 7

[1]

Both (7x + 1)° and (3x + 29)°

have the same intercepted arc.

So these two angles are congruent.

have the same intercepted arc.

So these two angles are congruent.

Close

## Inscribed Right Triangle

### Property

passes through the center of the circle.

### Example

The hypotenuse passes through

the center of the circle.

→ This is a right triangle.

→ Top angle: right angle

the center of the circle.

→ This is a right triangle.

→ Top angle: right angle

↓

x = 60 - [1]

Close

### Example

The hypotenuse passes through

the center of the circle.

→ This is a right triangle.

→ Top angle: right angle

the center of the circle.

→ This is a right triangle.

→ Top angle: right angle

↓

↓

x12 = 105 - [2]

x12 = 2

x = 24

[1]

(10, x, 26) triangle

→ ×2 of (5, 12, 13)

→ So draw (5, 12, 13) right triangle

next to the given triangle.

→ ×2 of (5, 12, 13)

→ So draw (5, 12, 13) right triangle

next to the given triangle.

[2]

Close

## Inscribed Quadrilateral

### Property

m∠1 + m∠2 = 180

m∠1' + m∠2' = 180

### Example

x + 100 = 180

x = 80

y + 70 = 180

y = 110

x = 80, y = 110

Close

## Tangent to a Circle

### Definition

that touches the circle.

From a point exterior of the circle,

you can draw two tangents.

### Property

are perpendicular

at the intersecting point of the circle.

### Example

### Property

### Example

Perimeter of △ABC

Solution P = 2(7 + 10 + 5)

= 2⋅22

= 44

Close

## Angle Formed by a Tangent and a Chord

### Formula

θ = 12mAB

### Example

mAB = 110, x = ?

Solution x = 12⋅110

= 55

Close

### Example

## Angle Formed by Two Chords

### Formula

θ = 12(mAC + mBD)

### Example

x = 12(32 + 74)

= 12⋅106

= 53

Close

## Angle Formed by Two Secants

### Formula

θ = 12(mBD - mAC)

### Example

x = 12(108 - 44)

= 12⋅64

= 32

Close

## Angle Formed by a Tangent and a Secant

### Formula

θ = 12(mBC - mAC)

### Example

x = 12(143 - 63)

= 12⋅80

= 40

Close

## Angle Formed by Two Tangents

### Formula

θ = 12(mABC - mAC)

### Example

α + 130 = 360

α = 230

↓

x = 12(230 - 130)

= 12⋅100

= 50

Close

## Segments Formed by Two Chords

### Formula

AP⋅PB = CP⋅PD

### Example

x⋅4 = 3⋅8

x = 3⋅2

= 6

Close

## Segments Formed by Two Secants

### Formula

PA⋅PB = PC⋅PD

### Example

x⋅(x + 3) = 4⋅(4 + 6)

x

^{2}+ 3x = 4⋅10

x

^{2}+ 3x = 40

x

^{2}+ 3x - 40 = 0

(x + 8)(x - 5) = 0 - [1] [2]

1) x + 8 = 0

x = -8( x ) - [3]

2) x - 5 = 0

x = 5

x = 5

[3]

x cannot be (-).

Close

## Segments Formed by a Tangent and a Secant

### Formula

PA

^{2}= PB⋅PC

### Example

x

^{2}= 4⋅(4 + 5)

= 4⋅9

= 36

x = 6 - [1]

[1]

x cannot be (-).

Close

## Area of a Circle

### Formula

A = πr

^{2}

### Example

Area?

Solution A = π⋅5

^{2}

= 25π

Close

### Example

## Area of the Sector of a Circle

### Formula

### Example

Area?

Solution A = π⋅6

^{2}⋅120360

= 36π⋅13

= 12π

Close