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.

#### Finding the Equation of a Circle

Step 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.

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

Step 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.

Let's 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.

Now, let's 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*}.

Finally, let's 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 to determine if the point \begin{align*}(14, 8)\end{align*} lies on the circle that is centered at the origin and has a diameter of 32 units.

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*} is 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.