##### Circle

### Definition

A circle is the set of all points in a plane that are equidistant from a given point that is called the origin.

Origin is the center of the circle

Radius is the distance from the center to any point on the circle. We label it with r.

Diameter is the segment that passes throw the origin and unites two points of the circle. We label it d and the diameter is twice the radius of the circle. d=2r

### Circumference of a circle

The circumference of a circle is the distance around the circle.The formula to determine the circumference is $ \displaystyle P=\pi d=2\pi r$. If we would form a circle with a twine than the circumference of the circle is the same as the length of the twine. If we know the perimeter of a circle then we can find the diameter and the radius respectively $ \displaystyle d=\frac{P}{\pi }$ and $ \displaystyle d=\frac{P}{\pi }$.

### Area of a Circle

The area of a circle is the region occupied by the circle in a two dimensional plane. The formula to determine the area is $ \displaystyle A=\pi {{r}^{2}}$. If we know the area of the circle we can find the radius by transforming the formula $ \displaystyle r=\sqrt{{\frac{A}{\pi }}}$

Example 1: Find the area and the circumference(perimeter) of a circle with a radius of 5 cm.

Solution: The area formula is: $ \displaystyle A=\pi {{r}^{2}}$

Substituting the radius we get:$ \displaystyle A=\pi {{(5)}^{2}}=25\pi c{{m}^{2}}$

Using $ \displaystyle \pi \approx 3,14$ we obtain: $ \displaystyle A=25\cdot 3,14=78,5 c{{m}^{2}}$

The circumference formula is: $ \displaystyle P=2\pi r$

Substituting the radius we get: $ \displaystyle P=2\pi \cdot 5=10\pi cm$

Using $ \displaystyle \pi \approx 3,14$ we obtain: $ \displaystyle P=10\pi =10\cdot 3,14=31,4cm$

Example 2: A circle has an area of $ \displaystyle A=16\pi {{m}^{2}}$.

a) Find the diameter of the circle.The area formula is $ \displaystyle A=\pi {{r}^{2}}$. To find the radius we transform our formula into $ \displaystyle r=\sqrt{{\frac{A}{\pi }}}$ and then calculate the radius of the circle $ \displaystyle r=\sqrt{{\frac{{16\pi }}{\pi }}}=\sqrt{{16}}=4m$.

We know that the diameter is twice of the radius so $ \displaystyle d=2r=2\cdot 4=8m$.

b)** Calculate the circumference of the circle.**

The circumference formula is:$ \displaystyle P=2\pi r$

Substituting the radius we found above we get: $ \displaystyle P=2\cdot 4\pi =8\pi m$

What is an Arc?

A chord separates the circumference of a circle into two sections the major arc and the minor arc.

So an arc is a part of the curve along the circumference of a circle.

**Arc Length** is the distance between two diffrent points on the circle or just a fraction of all the circumference of the entire circle.The formula to determine the lengh of an arc is $ \displaystyle l=\frac{n}{{{{{360}}^{\circ }}}}\times 2\pi r$ or $ \displaystyle l=\frac{n}{{{{{360}}^{\circ }}}}\times \pi d$ where “n“ is the angle measure.

Example 3: Calculate the length of this arc wich has a 50° angle.

Solution: The arc length formula is: $ \displaystyle l=\frac{n}{{{{{360}}^{\circ }}}}\times 2\pi r$

From the photo we know the angle and also the radius of the arc.

Substituting them we get:

$ \displaystyle l=\frac{{{{{50}}^{\circ }}}}{{{{{360}}^{\circ }}}}\times 2\pi \cdot 2$

$ \displaystyle =\frac{{{{5}^{\circ }}}}{{{{{36}}^{\circ }}}}\times 4\pi =1,74m$

So the length of our arc is: $ \displaystyle l=1,74m$

### What is a sector?

A sector is a portion of a circle resembling a pizza or the part of the circle area isolated by two radius and a circle arc.

The perimeter of a circle sector it consists of two radius and an arc.

Sector area is just a fraction of the area of the circle. The formula to determine the area of a sector is $ \displaystyle S=\frac{n}{{{{{360}}^{\circ }}}}\times \pi {{r}^{2}}$.

**Example 4**: Calculate the area of this sector which has a 120° angle.

**Solution: **The area of a sector formula is**: **$ \displaystyle S=\frac{n}{{{{{360}}^{\circ }}}}\times \pi {{r}^{2}}$

We know the angle and the radius of the sector and we substitute them at our formula.

$ \displaystyle S=\frac{{{{{120}}^{\circ }}}}{{{{{360}}^{{}^\circ }}}}\times \pi {{(4)}^{2}}$

$ \displaystyle =\frac{1}{3}\times 16\pi $

$ \displaystyle =8\pi =25,12c{{m}^{2}}$

So the area of our sector is: $ \displaystyle S=25,12c{{m}^{2}}$

### Inscribed and Central Angles

**Chord **is a line segment that touches 2 points on the circle.

An **inscribed angle** is an angle formed by two cords in a circle which have a common endpoint.

A **central angle** is an angle whose vertex is located at the center of a circle.

### Circle Theorems

Some theorems about circle are made from facts about angles on a circle.

**1.** The measure of an inscribed angle is half the size of a central angle with the same intercepted arc.

**2.** An angle inscribed across a circle diameter is always a right angle.

**3.** Two inscribed angles with the same intercepted arc are congruent.

**4.** Opposite angles in a cyclic quadrilateral add up to 180°.

### Some theorems about circle that are made from facts about radius,cords and tangents.

A tangent line to a circle is a line that touches the circle at exactly one point.

**1.** The tangent of a circle on a point it´s perpendicular with the radius at the same point.

**2.** The length of two tangents from a point to a circle are equal and the angles between the tangent and the radius are equal 90°.

**3.** Perpendicular form the center bisect the cord.

**4.** The angle between the tangent and the chord at the point of contact is equal to the angle in the alternate segment.

Example 5: Three points A, B and C line on the circumference of the circle where the line BC is the diameter and the angle C is given on the photo. Find the angle B.

Solution: We know that the sum of all angles on a triangle is 180°. We know only one angle of our triangle that is 60°.

Based on the circle theorems we know that:

“An angle inscribed across a circle diameter is always a right angle”

Now we know two angles of our triangle 60° and 90° ,so we find the missing one:

x+60°+90°=180°

x=180°-(60°+90°)

x=180°-150°=30°

So, our angle B is 30°.

Example 6: Find the missing angles.

Solution: Triangle ABC is an isosceles triangle based on the theorem “The length of two tangents from a point to a circle are equal”

Since$ \displaystyle AB=BC$ then also $ \displaystyle \widehat{{ABC}}=\widehat{{ACB}}$

We know that the sum of all angles on a triangle is 180°, from this we find our missing angle B.

$ \displaystyle \widehat{{ABC}}=\widehat{{ACB}}=x$

x+x+30°=180°

2x+30°=180°

2x=180°-30°

2x=150°

x=75°

From the theorem we also know that the angles between the tangent and the radius are equal and 90°.

So the angle $ \displaystyle \widehat{{ABO}}=\widehat{{ACO}}={{90}^{\circ }}$

Now from the fact that the sum of all angles on a quadrilateral is 360° we find our missing angle O.

30°+90°+90°+y=360°

y=360°-(90°+90°+30°)

y=360°-210°=150°

Based on the theorems we solved our problem and found our missing angles x=75° and y=150°.