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# 9.7: Extension: Writing and Graphing the Equations of Circles

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Graph a circle.
• Find the equation of a circle in the xy\begin{align*}x-y\end{align*} plane.
• Find the radius and center, given the equation of a circle and vice versa.
• Find the equation of a circle, given the center and a point on the circle.

## Graphing a Circle in the Coordinate Plane

Recall that the definition of a circle is the set of all points 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)\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, r\begin{align*}r\end{align*}. x\begin{align*}x\end{align*} is the horizontal distance y\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 x2+y2=r2\begin{align*}x^2+y^2=r^2\end{align*}.

Example 1: Graph x2+y2=9\begin{align*}x^2+y^2=9\end{align*}.

Solution: The center is (0, 0). It’s radius is the square root of 9, or 3. Plot the center, and then go out 3 units in every direction and connect them to form a circle.

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

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

If you square both sides of this equation, then we would have the standard equation of a circle.

Standard Equation of a Circle: The standard equation of a circle with center (h,k)\begin{align*}(h, k)\end{align*} and radius r\begin{align*}r\end{align*} is r2=(xh)2+(yk)2\begin{align*}r^2=(x-h)^2+(y-k)^2\end{align*}.

Example 2: Find the center and radius of the following circles.

a) (x3)2+(y1)2=25\begin{align*}(x-3)^2+(y-1)^2=25\end{align*}

b) (x+2)2+(y5)2=49\begin{align*}(x+2)^2+(y-5)^2=49\end{align*}

Solution:

a) Rewrite the equation as (x3)2+(y1)2=52\begin{align*}(x-3)^2+(y-1)^2=5^2\end{align*}. The center is (3, 1) and r=5\begin{align*}r = 5\end{align*}.

b) Rewrite the equation as (x(2))2+(y5)2=72\begin{align*}(x-(-2))^2+(y-5)^2=7^2\end{align*}. The center is (-2, 5) and r=7\begin{align*}r = 7\end{align*}.

When finding the center of a circle always take the opposite sign of what the value is in the equation.

Example 3: 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\begin{align*}r = 6\end{align*}. Plugging this into the equation of a circle, we get: (x(3))2+(y3)2=62\begin{align*}(x-(-3))^2+(y-3)^2=6^2\end{align*} or (x+3)2+(y3)2=36\begin{align*}(x+3)^2+(y-3)^2=36\end{align*}.

## Finding the Equation of a Circle

Example 4: Determine if the following points are on (x+1)2+(y5)2=50\begin{align*}(x+1)^2+(y-5)^2=50\end{align*}.

a) (8, -3)

b) (-2, -2)

Solution: Plug in the points for x\begin{align*}x\end{align*} and y\begin{align*}y\end{align*} in (x+1)2+(y5)2=50\begin{align*}(x+1)^2+(y-5)^2=50\end{align*}.

a) (8+1)2+(35)2=5092+(8)2=5081+6450\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+1)2+(25)2=50(1)2+(7)2=501+49=50\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

Example 5: Find the equation of the circle with center (4, -1) and passes through (-1, 2).

Solution: First plug in the center to the standard equation.

(x4)2+(y(1))2(x4)2+(y+1)2=r2=r2\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 x\begin{align*}x\end{align*} and y\begin{align*}y\end{align*} and solve for r\begin{align*}r\end{align*}.

(14)2+(2+1)2(5)2+(3)225+934=r2=r2=r2=r2\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 r2\begin{align*}r^2\end{align*}, the equation is (x4)2+(y+1)2=34\begin{align*}(x-4)^2+(y+1)^2=34\end{align*}.

## Review Questions

• Questions 1-4 are similar to Examples 1 and 2.
• Questions 5-8 are similar to Example 3.
• Questions 9-11 are similar to Example 4.
• Questions 12-15 are similar to Example 5.

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

1. (x+5)2+(y3)2=16\begin{align*}(x+5)^2+(y-3)^2=16\end{align*}
2. x2+(y+8)2=4\begin{align*}x^2+(y+8)^2=4\end{align*}
3. (x7)2+(y10)2=20\begin{align*}(x-7)^2+(y-10)^2=20\end{align*}
4. (x+2)2+y2=8\begin{align*}(x+2)^2+y^2=8\end{align*}

Find the equation of the circles below.

1. Is (-7, 3) on (x+1)2+(y6)2=45\begin{align*}(x+1)^2+(y-6)^2=45\end{align*}?
2. Is (9, -1) on (x2)2+(y2)2=60\begin{align*}(x-2)^2+(y-2)^2=60\end{align*}?
3. Is (-4, -3) on (x+3)2+(y3)2=37\begin{align*}(x+3)^2+(y-3)^2=37\end{align*}?
4. Is (5, -3) on (x+1)2+(y6)2=45\begin{align*}(x+1)^2+(y-6)^2=45\end{align*}?

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

1. center: (2, 3), point: (-4, -1)
2. center: (10, 0), point: (5, 2)
3. center: (-3, 8), point: (7, -2)
4. center: (6, -6), point: (-9, 4)

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