You draw a circle that is centered at the origin. You measure the diameter of the circle to be 32 units. Does the point \begin{align*}(14, 8)\end{align*} lie on the circle?

### Circles Centered at the Origin

Until now, your only reference to circles was from geometry. A **circle** is the set of points that are equidistant (the **radius**) from a given point (the **center**). A line segment that passes through the center and has endpoints on the circle is a **diameter**.

Now, we will take a circle and place it on the \begin{align*}x-y\end{align*} plane to see if we can find its equation. In this concept, we are going to place the center of the circle on the origin.

#### Investigation: Finding the Equation of a Circle

1. On a piece of graph paper, draw an \begin{align*}x-y\end{align*} plane. Using a compass, draw a circle, centered at the origin that has a radius of 5. Find the point \begin{align*}(3, 4)\end{align*} on the circle and draw a right triangle with the radius as the hypotenuse.

2. Using the length of each side of the right triangle, show that the Pythagorean Theorem is true.

3. Now, instead of using \begin{align*}(3, 4)\end{align*}, change the point to \begin{align*}(x, y)\end{align*} so that it represents any point on the circle. Using \begin{align*}r\end{align*} to represent the radius, rewrite the Pythagorean Theorem.

The **equation of a circle**, centered at the origin, is \begin{align*}x^2+y^2=r^2\end{align*}, where \begin{align*}r\end{align*} is the radius and \begin{align*}(x, y)\end{align*} is any point on the circle.

#### Solve the following problems

Find the radius of \begin{align*}x^2+y^2=16\end{align*} and graph.

To find the radius, we can set \begin{align*}16=r^2\end{align*}, making \begin{align*}r = 4\end{align*}. \begin{align*}r\end{align*} is not -4 because it is a distance and distances are always positive. To graph the circle, start at the origin and go out 4 units in each direction and connect.

Find the equation of the circle with center at the origin and passes through \begin{align*}(-7, -7)\end{align*}.

Using the equation of the circle, we have: \begin{align*}(-7)^2+(-7)^2=r^2\end{align*}. Solve for \begin{align*}r^2\end{align*}.

\begin{align*}(-7)^2+(-7)^2&=r^2 \\ 49+49&=r^2 \\ 98&=r^2\end{align*}

So, the equation is \begin{align*}x^2+y^2=98\end{align*}. The radius of the circle is \begin{align*}r=\sqrt{98}=7 \sqrt{2}\end{align*}.

Determine if the point \begin{align*}(9, -11)\end{align*} is on the circle \begin{align*}x^2+y^2=225\end{align*}.

Substitute the point in for \begin{align*}x\end{align*} and \begin{align*}y\end{align*} and see if it equals 225.

\begin{align*}9^2+(-11)^2&=225 \\ 81+121&\overset{?}{=}225 \\ 202& \ne 225\end{align*}

The point is not on the circle.

### Examples

#### Example 1

Earlier, you were asked does the point \begin{align*}(14, 8)\end{align*} lie on the circle.

From this lesson, you know that the equation of a circle that is centered at the origin is \begin{align*}x^2+y^2=r^2\end{align*}, where \begin{align*}r\end{align*} is the radius and \begin{align*}(x, y)\end{align*} is any point on the circle.

With the point \begin{align*}(14, 8)\end{align*}, \begin{align*}x = 14\end{align*} and \begin{align*}y = 8\end{align*}. We are given the diameter, but we need the radius. Recall that the radius is half the diameter, so the radius is \begin{align*}\frac{32}{2} = 16\end{align*}.

Plug these values into the equation of the circle. If they result in a true statement, the point lies on the circle.

\begin{align*}x^2+y^2=r^2\\ 14^2 + 8^2 \overset{?}{=} 16^2\\ 196 + 64 \overset{?}{=}256\\ 260 \ne 256\end{align*}

Therefore the point does not lie on the circle.

#### Example 2

Graph and find the radius of \begin{align*}x^2+y^2=4\end{align*}.

\begin{align*}r=\sqrt{4}=2\end{align*}

#### Example 3

Find the equation of the circle with a radius of \begin{align*}6 \sqrt{5}\end{align*}.

Plug in \begin{align*}6 \sqrt{5}\end{align*} for \begin{align*}r\end{align*} in \begin{align*}x^2+y^2=r^2\end{align*}

\begin{align*}x^2+y^2&=\left(6 \sqrt{5}\right)^2 \\ x^2+y^2&=6^2 \cdot \left(\sqrt{5}\right)^2 \\ x^2+y^2&=36 \cdot 5 \\ x^2+y^2&=180\end{align*}

#### Example 4

Find the equation of the circle that passes through \begin{align*}(5, 8)\end{align*}.

Plug in \begin{align*}(5, 8)\end{align*} for \begin{align*}x\end{align*} and \begin{align*}y\end{align*}, respectively.

\begin{align*}5^2+8^2&=r^2 \\ 25+64&=r^2 \\ 89&=r^2\end{align*}

The equation is \begin{align*}x^2+y^2=89\end{align*}

#### Example 5

Determine if \begin{align*}(-10, 7)\end{align*} in on the circle \begin{align*}x^2+y^2=149\end{align*}.

Plug in \begin{align*}(-10, 7)\end{align*} to see if it is a valid equation.

\begin{align*}(-10)^2+7^2&=149 \\ 100+49&=149\end{align*}

Yes, the point is on the circle.

### Review

Graph the following circles and find the radius.

- \begin{align*}x^2+y^2=9\end{align*}
- \begin{align*}x^2+y^2=64\end{align*}
- \begin{align*}x^2+y^2=8\end{align*}
- \begin{align*}x^2+y^2=50\end{align*}
- \begin{align*}2x^2+2y^2=162\end{align*}
- \begin{align*}5x^2+5y^2=150\end{align*}

Write the equation of the circle with the given radius and centered at the origin.

- 14
- 6
- \begin{align*}9 \sqrt{2}\end{align*}

Write the equation of the circle that passes through the given point and is centered at the origin.

- \begin{align*}(7,-24)\end{align*}
- \begin{align*}(2,2)\end{align*}
- \begin{align*}(-9,-10)\end{align*}

Determine if the following points are on the circle, \begin{align*}x^2+y^2=74\end{align*}.

- \begin{align*}(-8,0)\end{align*}
- \begin{align*}(7,-5)\end{align*}
- \begin{align*}(6,-6)\end{align*}

** Challenge** In Geometry, you learned about tangent lines to a circle. Recall that the tangent line touches a circle at one point and is perpendicular to the radius at that point, called the point of tangency.

- The equation of a circle is \begin{align*}x^2+y^2=10\end{align*} with point of tangency \begin{align*}(-3, 1)\end{align*}.
- Repeat the steps in #16 to find the equation of the tangent line to \begin{align*}x^2+y^2=34\end{align*} with a point of tangency of \begin{align*}(3, 5)\end{align*}.

### Answers for Review Problems

To see the Review answers, open this PDF file and look for section 10.3.