## The definition and basics of a circle

Let $S$ be a fixed point in the plane and $r>0$. **A circle** is a set of all points of a plane which are equidistant from point $S$ which is equal to $r.$ Point $S$ is called the* center of the circle*. The distance $r$ from the center to any point of the circle is called the *radius of the circle*. A circle with a center situated in point $S$ and with radius $r$ is denoted by $c(S,r)$.

**A chord** is the line segment in which the ends are two different points that lie on the circumference of a circle. A chord is denoted by $c$.

The longest chord is called the **diameter**. A diameter contains the center of the circle. The length of a diameter is equal to $2r.$ A diameter is denoted by $R$.

**A sector** is an enclosed part of a circle which contains two radii and a part of a circumference.

**A segment*** *is a part of a circle enclosed by a chord and a part of the circumference.

**A secant** of a circle is the line that intersects a circle in two points and **a tangent line **to the circle is a line that has exactly one point in common with the circle. A secant is denoted by $s$ and a tangent by $t$.

## A secant and tangent to the circle

Let $t$ be the tangent to the circle $c(S,|SD|)$ and a tangent with the circle has one in common point $D$. Point $D$ is called** a point of tangency **and is valid: $\overline{SD} \perp t$.

**Concentric circles** have the same center but different radius.

If $r$ is the length of the circle, then its circumference is given with the formula:

*$$ C = 2 r \pi,$$*

and area with

*$$A = r^2 \pi.$$
*

From this we can see that the number $\pi$ is defined as the ration of the length of circumference and its diameter.

## Length of the arc and area of a sector

Firstly, consider circle on which lie two different points $A$ and $B$. Those points divide it in two different arcs. Their union is the whole circle.

For denoting arcs we must refer to the positive direction – in an anticlockwise motion.

This means that $\widehat{AB} \not= \widehat{BA}$.

We can draw an angle whose axis is the center of the circle and whose legs intersect the circle at points $A$ and $B$. This angle formed is called **the central angle**.

A whole circle has a circumference of $ 360^{\circ}$. This means that we can make the following ratio:

$$ l ( 1^{\circ}) = \frac{2r\pi}{360^{\circ}}$$

i.e.

$$ l ( 1^{\circ}) = \frac{r\pi}{180^{\circ}},$$

where $l$ is length of the arc.

If the central angle has $\alpha$ degrees; than the length of the arc is $\alpha$ times greater than the arc that matches the $ 1^{\circ}$ angle:

$$ l (\alpha) = \frac{r\pi\alpha}{180^{\circ}}.$$

In a similar way we can find the formula for the area of a sector:

$$ P (1^{\circ}) = \frac{r^2 \pi}{360^{\circ}}$$

and

$$ P (\alpha) = \frac{r^2 \pi\alpha}{360^{\circ}}.$$

## Central and inscribed angles

Every angle whose vertex is a point on the circle and whose legs intersect the circle is called an **inscribed angle.**

If we mark the points in which the legs of the inscribed angle intersect the circle with $A$ and $B$, then we call that angle the inscribed angle over the segment $AB$.

For these two points we can also define another angle whose legs also run through those points, but its vertex is placed in the center. This is called the **central angle**.

**The central and inscribed angle theorem**

The central angle over the arc of the circle is equal to the double inscribed angle over that same arc.

*Proof*. There are exactly three different cases that may appear.

1.) The* c*enter of the circle lies on one leg of the inscribed angle.

In this case we can observe triangle $TSB$. We have:

$$\mid TS \mid = \mid SB \mid = r,$$

which means that triangle $TSB$ is an isosceles triangle which implies $STB=SBT=\beta$. $\alpha$ is the external angle of triangle $TSB$ which means that the measure of angle $\alpha$ is equal to the sum of measures of angles $STB$ and $SBT$, which is equal to $2\beta$.

2.) The center of the circle lies within inscribed angle.

First of all, determine point $C$ in the arc $\widehat{AB}$ which is diametrically opposite of point $T$. We then apply case 1.) on inscribed angles $\angle{ATS}$ and $\angle{BTS}$:

$\angle{ASC}+\angle{BSC}=2\angle{ATS}+2\angle{BTS}=2(\angle{ATS}+\angle{BTS}).$

Considering that

$\beta = \angle{ATS}+ \angle{BTS}$ and $\alpha =\angle{ASC}+\angle{BSC}$, we have:

$\alpha = 2 \beta.$

3.) The center lies outside of the inscribed circle.

To prove this case we will also use case 1.). First, we have that $\alpha =\angle{ASB}= \angle{ASC} – \angle{CSB}$ and $\beta = \angle{ATB}=\angle{ATC} – \angle{BTC}$. Then we have:

$ \angle{ASC} = 2 \angle{ATC}$,

$\angle{BSC} = 2 \angle{BTC}$.

.When we subtract these two equations we get:

$\angle{ASC} – \angle{BSC} = 2 (\angle{ATC} -\angle{BTC} )$.

It follows:

$\alpha = 2\beta$.

*Example 1.* Find the missing angle:

*Solution:*

$\alpha = 2 \beta = 2 \cdot 50^{\circ} = 100^{\circ}.$

**Thales’ theorem**

Every inscribed angle over diameter of the circle is the right angle.

*Proof.*

First, draw a circle and a triangle whose one side is the diameter and whose vertex is any point $C$ on the circle different from $A$ and $B$.

Draw a segment line that connects point $C$ and the center of the circle $S$.

Triangle $BSC$ is an isosceles triangle because $|BS|=|CS|$ (length of $\overline{BS}$ and $\overline{CS}$ is equal to the radius of the circle). It follows $\angle SBC = \angle SCB = \alpha$.

For the same reason triangle $ASC$ is an isosceles triangle and $\angle SAC = \angle SCA = \beta$.

Now, we have:

$ 2 \alpha + 2 \beta = 180^{\circ} \Rightarrow \alpha + \beta = 90^{\circ}.$

**Reverse of Thales’ theorem**

Center of the circumscribed circle of right triangle is located in the middle of hypotenuse.

Divide the angle at point $C$ with line $CX$ in a way that $\angle ACX = \alpha$.

Triangle $ACX$ is an isosceles triangle, it follows $\mid AX\mid = \mid CX \mid$ and $\angle ACX = \angle CAX = \alpha$.

Also, triangle $CBX$ is an isosceles triangle, it follows $\mid BX\mid = \mid CX \mid$ and $\angle BCX = \angle CBX = \beta$.

Thus we obtain $\mid BX \mid = \mid CX \mid = \mid AX \mid.$

This means that $X$ is the center of the circumscribed circle and the middle point of the hypotenuse.

## Equation of a circle

Every curve has an equation. From this equation we can find all the information we require to draw the curve and know each one of its properties.

A circle’s equation can be derived from its definition.

Let’s say that the coordinate of the circles center is $S(p, q)$ and any point $T$ on the circumference of a circle has the coordinate $T(x, y)$. From the formula for the distance of two points and the fact that we know that the distance is equal to the length of the radius $r$ we get that:

$$\sqrt{(x – p)^2 + (y – q)^2} = r$$

i.e.

$$\ (x – p)^2 + (y – q)^2 = r^2.$$

If $S(p, q)$ is the center of the circle, and $r$ its radius, then the equation of a circle is:

$$\ (x – p)^2 + (y – q)^2 = r^2.$$

What if we put the center of the circle in origin and set the radius of $1$? We get a unit circle.

Its equation is: $ x^2 + y^2 = 1.$

*Example 2.* Draw a circle that is described by the following equation:

$$ (x – 2)^2 + (y + 2)^2 = 9.$$

*Solution*:

From the left side of this equation we can find the center of the circle. The $x$ coordinate of the center is $2$, and the $y$ coordinate is $- 2$.

From the right side of the equation we can find radius. Radius is equal to $\sqrt{9} = \pm 3$. Since a radius can only be a positive number we can see that the radius is equal to $3$.

## The mutual position of a line and a circle

A line and the circle can be in three positions:

1. Two different points – line is the secant of the circle.

2. One point in common – line is a tangent to the circle.

3. No points in common

How can you find out in which relation are the circle and the line without having to draw them? You can do that using their equations.

To find their relation we have to find their common points.

*Example 3*. In which relation are the line $2x – y + 3 = 0$ and the circle $ x^2 + y^2 – 7x + 2y – 53 = 0$?

*Solution:*

First, from the line equation we extract $y$ or $x$. in this example we will extract $y$:

$y = 2x + 3.$

And then insert this $y$ into the equation.

$ x^2 + (2x + 3)^2 – 7x + 2 (2x + 3) – 53 = 0$

$\Rightarrow x^2 + 4x^2 + 12x + 9 – 7x + 4x + 6 – 53 = 0$

$\Rightarrow 5x^2 + 9x – 38 = 0.$

By solving this quadratic equation we get the solutions $ x_1 = 2$, $ x_2 = \frac{16}{5}$. By inserting them into any of these equations we get matching $y$ coordinates and finally the points $ (2, 7), (\frac{16}{5}, \frac{23}{5})$. This means that the given line and circle intersect in two points.

If we got only one point this would mean that the line is a tangent to the circle and if the solutions were complex this would mean that the line and the circle have no points in common.

**Power of the point considering circle**

Let $c(S,r)$ be a circle and $T$ any point of the plane. For any line $p$ which goes through point $T$, the product is a constant, where $A$ and $B$ are the intersections of the line $p$ and circle $c$. I.e.:

$$|TA|\cdot|TB|=|ST|^{2} – r^{2}.$$

Real number $|TA| \cdot |TB|$ is called **power of the point $T$ considering circle $c$**.

Depending on whether the point $T$ is located; $T$ is on the circumference of the circle $c(S, r)$, $T$ lies within the circle $c(S, r)$ or $T$ is outside of the circle $c(S, r)$, the power of the point $T$ relative to the circle is equal to $0$, negative or positive.

## A tangent and normal to the circle

The equation of a tangent of the circle $ (x – p)^2 + (y + q)^2 = r^2$ with touching point $ T_0 (x_0, y_0)$ is:

$$ (x_0 – p )(x – p) + (y_0 – q)(y – q) = r^2.$$

**A normal** to the circle is a line that is perpendicular to the tangent and goes through the touching point of the tangent and the circle.

The equation of a normal to the circle $ (x – p)^2 + (y + q)^2 = r^2$ with touching point of the tangent and the circle $ T_0 (x_0, y_0)$ is:

$$ y – y_0 = \frac{y_0 – q}{x_0 – p} (x – x_0).$$

*Example 4*. Determine the equations of tangent and normal of a circle $ x^2 + y^2 + 8x – 6y – 1 = 0$ in the point $T$ with coordinates $T (1, 4)$.

*Solution*:

First thing we have to do is convert this equation to a form we are used to:

$$ (x^2 + 8x) + (y^2 – 6y) – 1 = 0\Rightarrow (x + 4)^2 – 16 + (x – 3)^2 – 9 – 1 = 0 \Rightarrow (x + 4)^2 + (x – 3)^2 = 26. $$

From the equation of a circle we can see that the center of the circle is located in point $S( -4, 3)$.

Points $S$ and $T$ determine a normal to the circle, so we get that

$$y-3=\frac{4-3}{-4-1} (x+4),$$

i.e.

$$y=-\frac{1}{5} x + \frac{11}{5}$$

is the required equation of a normal to the circle $(x + 4)^2 + (x – 3)^2 = 26$ in point $T$. The equation of a tangent to the circle we will base using the perpendicularity clause of two lines. We get that

$$y=5x + \frac {11}{5}$$

is the required equation of a tangent to the circle $(x + 4)^2 + (x – 3)^2 = 26$ in point $T$.

## Circle worksheets

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