# 6.21: Circles in the Coordinate Plane

- Page ID
- 5046

Graph a circle. Use \((h, k)\) as the center and a point on the circle. Formula: \((x-h)^2 + (y-k)^2 = r^2\) where \((h, k)\) is the center and \(r\) is the radius.

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 \((x,y)\) is a point on the circle, then the distance from the center to this point would be the **radius**, r. x is the horizontal distance and y is the vertical distance. This forms a right triangle. From the Pythagorean Theorem, the equation of a circle * centered at the origin* is \(x^2+y^2=r^2\).

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

\(r=\sqrt{(x−h)^2+(y−k)^2}\)

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 \((h,k)\) and radius \(r\) is \(r^2=(x−h)^2+(y−k)^2\).

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?

Example \(\PageIndex{1}\)

Find the center and radius of the following circle.

\((x+2)^2+(y−5)^2=49\)

**Solution**

Rewrite the equation as \((x−(−2))^2+(y−5)^2=7^2\). The center is \((-2, 5)\) and \(r=7\).

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 \(\PageIndex{2}\)

Find the center and radius of the following circle.

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

**Solution**

First plug in the center to the standard equation.

\(\begin{aligned} (x−4)^2+(y−(−1))^2&=r^2 \\ (x−4)^2+(y+1)^2&=r^2\end{aligned}\)

Now, plug in (-1, 2) for \(x\) and \(y\) and solve for \(r\).

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

Substituting in \(34\) for \(r^2\), the equation is \((x−4)^2+(y+1)^2=34\).

Example \(\PageIndex{3}\)

Graph \(x^2+y^2=9\).

**Solution**

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.

Example \(\PageIndex{4\)

Find the equation of the circle below.

**Solution**

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 \(r=6\). Plugging this into the equation of a circle, we get: \((x−(−3))^2+(y−3)^2=6^2\) or \((x+3)^2+(y−3)^2=36\).

Example \(\PageIndex{5}\)

Determine if the following points are on \((x+1)^2+(y−5)^2=50\).

**Solution**

Plug in the points for x and y in \((x+1)^2+(y−5)^2=50\).

- \((8, -3)\)

\(\begin{aligned} (8+1)^2+(−3−5)^2&=50 \\ 9^2+(−8)^2&=50 \\ 81+64 &\neq 50\end{aligned}\)

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

- \((-2, -2)\)

\(\begin{aligned} (−2+1)^2+(−2−5)^2&=50 \\ (−1)^2+(−7)^2&=50 \\ 1+49&=50\end{aligned}\)

\( (-2, -2)\) is on the circle

## Review

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

- \((x+5)^2+(y−3)^2=16\)
- \(x^2+(y+8)^2=4
- \((x−7)^2+(y−10)^2=20
- \((x+2)^2+y^2=8

Find the equation of the circles below.

- Is (-7, 3) on \((x+1)^2+(y−6)^2=45\)?
- Is (9, -1) on \((x−2)^2+(y−2)^2=60\)?
- Is (-4, -3) on \((x+3)^2+(y−3)^2=37\)?
- Is (5, -3) on \((x+1)^2+(y−6)^2=45\)?

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)

## Review (Answers)

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

## Vocabulary

Term | Definition |
---|---|

circle |
The set of all points that are the same distance away from a specific point, called the .center |

radius |
The distance from the center to the outer rim of a circle. |

Distance Formula |
The distance between two points \((x_1, y_1)\) and \((x_2, y_2)\) can be defined as \(d=\sqrt{(x_2−x_1)^2+(y_2−y_1)^2}\). |

Origin |
The origin is the point of intersection of the x and y axes on the Cartesian plane. The coordinates of the origin are (0, 0). |

## Additional Resources

Interactive Element

Video: Graphing Circles Principles - Basic

Activities: Circles in the Coordinate Plane Discussion Questions

Study Aids: Properties of a Circle Study Guide

Practice: Circles in the Coordinate Plane

Real World: GPS Geometry