What if you were given the length of the radius of a circle and the coordinates of its center? How could you write the equation of the circle in the coordinate plane?

### Circles in the Coordinate Plane

Recall that a circle is the set of all points in a plane that are the same distance from the center. This definition can be used to find an equation of a circle in the coordinate plane.

Let’s start with the circle centered at (0, 0). If \begin{align*}(x, y)\end{align*} is a point on the circle, then the distance from the center to this point would be the radius, \begin{align*}r\end{align*}. \begin{align*}x\end{align*} is the horizontal distance and \begin{align*}y\end{align*} is the vertical distance. This forms a right triangle. From the Pythagorean Theorem, the equation of a circle *centered at the origin* is \begin{align*}x^2+y^2=r^2\end{align*}.

The center does not always have to be on (0, 0). If it is not, then we label the center \begin{align*}(h, k)\end{align*}. We would then use the Distance Formula to find the length of the radius.

\begin{align*}r=\sqrt{(x-h)^2+(y-k)^2}\end{align*}

If you square both sides of this equation, then you would have the standard equation of a circle. **The standard equation of a circle with center \begin{align*}(h, k)\end{align*} and radius \begin{align*}r\end{align*} is \begin{align*}r^2=(x-h)^2+(y-k)^2\end{align*}.**

#### Graphing a Circle

Graph \begin{align*}x^2+y^2=9\end{align*}.

The center is (0, 0). Its radius is the square root of 9, or 3. Plot the center, plot the points that are 3 units to the right, left, up, and down from the center and then connect these four points to form a circle.

#### Finding the Equation of a Circle

Find the equation of the circle below.

First locate the center. Draw in the horizontal and vertical diameters to see where they intersect.

From this, we see that the center is (-3, 3). If we count the units from the center to the circle on either of these diameters, we find \begin{align*}r = 6\end{align*}. Plugging this into the equation of a circle, we get: \begin{align*}(x-(-3))^2+(y-3)^2=6^2\end{align*} or \begin{align*}(x+3)^2+(y-3)^2=36\end{align*}.

#### Determining if Points are on a Circle

Determine if the following points are on \begin{align*}(x+1)^2+(y-5)^2=50\end{align*}.

Plug in the points for \begin{align*}x\end{align*} and \begin{align*}y\end{align*} in \begin{align*}(x+1)^2+(y-5)^2=50\end{align*}

a) (8, -3)

\begin{align*}(8+1)^2+(-3-5)^2 &= 50\\ 9^2+(-8)^2 &= 50\\ 81+64 & \ne 50\end{align*}

(8,-3) is *not* on the circle.

b) (-2, -2)

\begin{align*}(-2+1)^2+(-2-5)^2 &= 50\\
(-1)^2+(-7)^2 &= 50\!\\
1+49 &= 50\end{align*}

(-2, -2) is on the circle

### Examples

Find the center and radius of the following circles.

#### Example 1

\begin{align*}(x-3)^2+(y-1)^2=25\end{align*}

Rewrite the equation as \begin{align*}(x-3)^2+(y-1)^2=5^2\end{align*}. The center is (3, 1) and \begin{align*}r = 5\end{align*}.

#### Example 2

\begin{align*}(x+2)^2+(y-5)^2=49\end{align*}

Rewrite the equation as \begin{align*}(x-(-2))^2+(y-5)^2=7^2\end{align*}. The center is (-2, 5) and \begin{align*}r = 7\end{align*}.

Keep in mind that, due to the minus signs in the formula, the coordinates of the center have the ** opposite signs** of what they may initially appear to be.

#### Example 3

Find the equation of the circle with center (4, -1) and which passes through (-1, 2).

First plug in the center to the standard equation.

\begin{align*}(x-4)^2+(y-(-1))^2&=r^2 \\ (x-4)^2+(y+1)^2&=r^2\end{align*}

Now, plug in (-1, 2) for \begin{align*}x\end{align*} and \begin{align*}y\end{align*} and solve for \begin{align*}r\end{align*}.

\begin{align*}(-1-4)^2+(2+1)^2&=r^2\\ (-5)^2+(3)^2&=r^2\\ 25+9&=r^2\\ 34&=r^2\end{align*}

Substituting in 34 for \begin{align*}r^2\end{align*}, the equation is \begin{align*}(x-4)^2+(y+1)^2=34\end{align*}.

### Interactive Practice

### Review

Find the center and radius of each circle. Then, graph each circle.

- \begin{align*}(x+5)^2+(y-3)^2=16\end{align*}
- \begin{align*}x^2+(y+8)^2=4\end{align*}
- \begin{align*}(x-7)^2+(y-10)^2=20\end{align*}
- \begin{align*}(x+2)^2+y^2=8\end{align*}

Find the equation of the circles below.

- Determine if the following points are on \begin{align*}(x+1)^2+(y-6)^2=45\end{align*}.
- (2, 0)
- (-3, 4)
- (-7, 3)

Find the equation of the circle with the given center and point on the circle.

- center: (2, 3), point: (-4, -1)
- center: (10, 0), point: (5, 2)
- center: (-3, 8), point: (7, -2)
- center: (6, -6), point: (-9, 4)
- Now let’s find the equation of a circle using three points on the circle. Given the points \begin{align*}A(-12, -21), B(2, 27)\end{align*} and \begin{align*}C(19, 10)\end{align*} on the circle (an arc could be drawn through these points from \begin{align*}A\end{align*} to \begin{align*}C\end{align*}), follow the steps below.
- Since the perpendicular bisector passes through the midpoint of a segment we must first find the midpoint between \begin{align*}A\end{align*} and \begin{align*}C\end{align*}.
- Now the perpendicular line must have a slope that is the opposite reciprocal of the slope of \begin{align*}\overleftrightarrow{AC}\end{align*}. Find the slope of \begin{align*}\overleftrightarrow{AC}\end{align*} and then its opposite reciprocal.
- Finally, you can write the equation of the perpendicular bisector of \begin{align*}\overline{AC}\end{align*} using the point you found in part a and the slope you found in part b.
- Repeat steps a-c for chord \begin{align*}\overline{BC}\end{align*}.
- Now that we have the two perpendicular bisectors of the chord we can find their intersection. Solve the system of linear equations to find the center of the circle.
- Find the radius of the circle by finding the distance from the center (point found in part \begin{align*}e\end{align*}) to any of the three given points on the circle.
- Now, use the center and radius to write the equation of the circle.

Find the equations of the circles which contain the three points.

- \begin{align*}A(-2, 5), B(5, 6)\end{align*} and \begin{align*}C(6, -1)\end{align*}
- \begin{align*}A(-11, -14), B(5, 16)\end{align*} and \begin{align*}C(12, 9)\end{align*}

### Review (Answers)

To view the Review answers, open this PDF file and look for section 9.12.