9.12: Circles in the Coordinate Plane
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? After completing this Concept, you'll be able to write the standard equation of a circle.
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CK12 Foundation: Chapter9CirclesintheCoordinatePlaneA
James Sousa: Write the Standard Form of a Circle
Guidance
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*}
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*}
\begin{align*}r=\sqrt{(xh)^2+(yk)^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*}
Example A
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.
Example B
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*}
Example C
Determine if the following points are on \begin{align*}(x+1)^2+(y5)^2=50\end{align*}
a) (8, 3)
b) (2, 2)
Plug in the points for \begin{align*}x\end{align*}
a) \begin{align*}(8+1)^2+(35)^2 &= 50\\
9^2+(8)^2 &= 50\\
81+64 & \ne 50\end{align*}
(8, 3) is not on the circle
b) \begin{align*}(2+1)^2+(25)^2 &= 50\\
(1)^2+(7)^2 &= 50\!\\
1+49 &= 50\end{align*}
(2, 2) is on the circle
Watch this video for help with the Examples above.
CK12 Foundation: Chapter9CirclesintheCoordinatePlaneB
Vocabulary
A circle is the set of all points that are the same distance away from a specific point, called the center. A radius is the distance from the center to the circle.
Guided Practice
Find the center and radius of the following circles.

\begin{align*}(x3)^2+(y1)^2=25\end{align*}
(x−3)2+(y−1)2=25 
\begin{align*}(x+2)^2+(y5)^2=49\end{align*}
(x+2)2+(y−5)2=49
3. Find the equation of the circle with center (4, 1) and which passes through (1, 2).
Answers:
1. Rewrite the equation as \begin{align*}(x3)^2+(y1)^2=5^2\end{align*}
2. Rewrite the equation as \begin{align*}(x(2))^2+(y5)^2=7^2\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.
3. First plug in the center to the standard equation.
\begin{align*}(x4)^2+(y(1))^2&=r^2 \\
(x4)^2+(y+1)^2&=r^2\end{align*}
Now, plug in (1, 2) for \begin{align*}x\end{align*}
\begin{align*}(14)^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*}
Interactive Practice
Practice
Find the center and radius of each circle. Then, graph each circle.

\begin{align*}(x+5)^2+(y3)^2=16\end{align*}
(x+5)2+(y−3)2=16 
\begin{align*}x^2+(y+8)^2=4\end{align*}
x2+(y+8)2=4 
\begin{align*}(x7)^2+(y10)^2=20\end{align*}
(x−7)2+(y−10)2=20 
\begin{align*}(x+2)^2+y^2=8\end{align*}
(x+2)2+y2=8
Find the equation of the circles below.
 Determine if the following points are on \begin{align*}(x+1)^2+(y6)^2=45\end{align*}
(x+1)2+(y−6)2=45 . (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*}
A(−12,−21),B(2,27) and \begin{align*}C(19, 10)\end{align*}C(19,10) on the circle (an arc could be drawn through these points from \begin{align*}A\end{align*}A to \begin{align*}C\end{align*}C ), 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*}
A 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 ac 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.
 Since the perpendicular bisector passes through the midpoint of a segment we must first find the midpoint between \begin{align*}A\end{align*}
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*}
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Image Attributions
Here you'll learn how to find the standard equation for circles given their radius and center. You'll also graph circles in the coordinate plane.