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

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## Learning Objectives

• Graph a circle.
• Find the equation of a circle in the $x-y$ 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 $(x, y)$ 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 $y$ is the vertical distance. This forms a right triangle. From the Pythagorean Theorem, the equation of a circle, centered at the origin is $x^2+y^2=r^2$.

Example 1: Graph $x^2+y^2=9$.

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 $(h, k)$ and would 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 we would have the standard equation of a circle.

Standard Equation of a Circle: The standard equation of a circle with center $(h, k)$ and radius $r$ is $r^2=(x-h)^2+(y-k)^2$.

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

a) $(x-3)^2+(y-1)^2=25$

b) $(x+2)^2+(y-5)^2=49$

Solution:

a) Rewrite the equation as $(x-3)^2+(y-1)^2=5^2$. The center is (3, 1) and $r = 5$.

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

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

## Finding the Equation of a Circle

Example 4: Determine if the following points are on $(x+1)^2+(y-5)^2=50$.

a) (8, -3)

b) (-2, -2)

Solution: Plug in the points for $x$ and $y$ in $(x+1)^2+(y-5)^2=50$.

a) $(8+1)^2+(-3-5)^2=50\!\\9^2+(-8)^2=50\!\\81+64 \ne 50$

(8, -3) is not on the circle

b) $(-2+1)^2+(-2-5)^2=50\!\\(-1)^2+(-7)^2=50\!\\1+49=50$

(-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.

$(x-4)^2+(y-(-1))^2&=r^2 \\(x-4)^2+(y+1)^2&=r^2$

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

$(-1-4)^2+(2+1)^2&=r^2\\(-5)^2+(3)^2&=r^2\\25+9&=r^2\\34&=r^2$

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

## 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. $(x+5)^2+(y-3)^2=16$
2. $x^2+(y+8)^2=4$
3. $(x-7)^2+(y-10)^2=20$
4. $(x+2)^2+y^2=8$

Find the equation of the circles below.

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

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)

8 , 9 , 10

Feb 22, 2012

Aug 21, 2014