The following are short descriptions of the circle shown below.

**Tangent**

*Tangent*is a line that would pass through one point on the circle.

**Secant**

*Secant*is a line that would pass through two points on the circle.

**Chord**

*Chord*is a secant that would terminate on the circle itself.

**Diameter,**

*d**Diameter*is a chord that passes through the center of the circle.

**Radius,**

*r**Radius*is one-half of the diameter.

**Area of the circle**- $A = \pi \, r^2$
$A = \frac{1}{4}\pi \, d^2$

**Circumference of the circle**- $c = 2\pi \, r$
$c = \pi \, d$

**Sector of a Circle**- Length of arc:
$s = \dfrac{\pi \, r \theta_{degree}}{180^\circ}$

$s = r \, \theta_{radians}$

Area of the sector:

$A = \dfrac{\pi \, r^2 \theta_{degrees}}{360^\circ}$$A = \frac{1}{2}r^2 \, \theta_{radians} $

$A = \frac{1}{2}sr$

**Segment of a Circle**- Area of circular segment with
*s*< ½*c*:

$A = A_{sector} - A_{triangle}$$A = \frac{1}{2}r^2 (\theta_{radian} - \sin \theta_{degrees})$

Area of circular segment with

*s*> ½*c*:

$A = A_{sector} + A_{triangle}$$A = \frac{1}{2}r^2 (\alpha_{radian} + \sin \theta_{degrees})$

## Relationship Between Central Angle and Inscribed Angle

**Central angle**

**Inscribed angle**

*circumferential angle*and

*peripheral angle*.

Figure below shows a central angle and inscribed angle intercepting the same arc AB. The relationship between the two is given by

if and only if both angles intercepted the same arc. In the figure below, θ and α intercepted the same arc AB.

Watch our video below for proof of this relationship.

## Applications of the Relationship Between Central Angle and Inscribed Angle

**Right Triangle Inscribed in a Circle**

We can also say that an angle inscribed in a semicircle is a right angle. From the figure above, the diameter AC is the hypotenuse of triangles AB_{1}C, AB_{2}C, AB_{3}C, and AB_{4}C.

**Intersecting Chords**

$\dfrac{opposite\,\,to\,\,\theta}{opposite\,\,to\,\,\beta} = \dfrac{DE}{AE} = \dfrac{CE}{BE}$

$BE \times DE = AE \times CE$

This means that for intersecting chords in a circle, the product of segments of one is equal to the product of segments of the other.

**Intersecting Secants**

Therefore, triangles EAC and EBD are similar, and by ratio and proportion of similar triangles

$\dfrac{opposite\,\,to\,\,\phi}{opposite\,\,to\,\,\beta} = \dfrac{DE}{BE} = \dfrac{CE}{AE}$

$DE \times AE = CE \times BE$

Also note that

$\varphi = \frac{1}{2}(\theta - \beta)$

**Intersecting Tangent and Secant**

Triangle ABE is similar to triangle BCE. By ratio and proportion

$\dfrac{opposite\,\,to\,\,\phi}{opposite\,\,to\,\,\beta} = \dfrac{BE}{AE} = \dfrac{CE}{BE}$

$BE^2 = CE \times AE$