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# Equations of Circles

## (x - h)^2 + (y-k)^2 = r^2

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Equations of Circles

What type of graph is produced by the equation $x^2-2x+y^2-4y+1=0$ ? How can you turn this equation into graphing form in order to graph it?

#### Guidance

A circle is one example of a conic section. A circle is the set of all points that are equidistant from a center point. The general equation of a circle is  $(x-h)^2+(y-k)^2=r^2$ where  $(h, k)$ is the center of the circle and  $r$ is the radius of the circle.

You can derive the equation of a circle with the help of the Pythagorean Theorem.

Example A

Label the three sides of the right triangle below in terms of $x, y$ , and $r$ . Then, use the Pythagorean Theorem to write an equation that shows the relationship between the three sides.

Solution: The three sides of the triangle are $x$ , $y$ , and $r$ , where  $r$ is the hypotenuse.

By the Pythagorean Theorem, $x^2+y^2=r^2$ .

Example B

Do all points on the circle from Example A satisfy the equation $x^2+y^2=r^2$ ?

Solution: Yes, anywhere you move the point $(x, y)$ , the lengths of the sides of the triangle will still be  $x$$y$ , and $r$ , so  $x^2+y^2=r^2$ will still be true.

In fact, the set of points that make up the circle is exactly the set of points that satisfy the equation $x^2+y^2=r^2$ . This is why the general equation of a circle centered at the origin is $x^2+y^2=r^2$ , where  $r$ is the radius of the circle.

Example C

Why does a circle with center  $(h, k)$ and radius  $r$ have the general equation $(x-h)^2+(y-k)^2=r^2$ ?

Solution: Pick a point on the circle and draw a right triangle connecting that point with the center of the circle.

The three sides of the triangle have lengths $x-h$$y-k$ and $r$ .

The relationship between the sides of the triangle is shown with the Pythagorean Theorem:

$(x-h)^2+(y-k)^2=r^2$

Points on the other side of the circle produce triangles with slightly different side lengths:

However, the relationship between the three sides of this triangle is the same, as shown below:

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

The set of points that satisfy the equation  $(x-h)^2+(y-k)^2=r^2$ are the set of points that make up the circle.

Concept Problem Revisited

The equation  $x^2-2x+y^2-4y+1=0$ is the equation for a circle. You can tell it is a circle because there is both an  $x^2$ term and a  $y^2$ term, and the coefficients of each are the same. To turn this equation into the form $(x-h)^2+(y-k)^2=r^2$ , you must complete the square twice. Remember that when completing the square, you are trying to figure out what number to add to make a perfect square trinomial. In order to maintain the equality of the equation, you must add the same numbers to both sides of the equation.

$& x^2-2x+ \underline{\; \; \; \; \; \; \; \;}+y^2-4y+\underline{\; \; \; \; \; \; \; \;}=-1 \\& x^2-2x+ \underline{1} +y^2-4y+ \underline{4} =-1 \underline{+1+4} \\& x^2-2x+1+y^2-4y+4=4 \\& (x-1)^2+(y-2)^2=4$

#### Vocabulary

A circle is the set of all points equidistant from a given point. The general equation of a circle is $(x-h)^2+(y-k)^2=r^2$ .

#### Guided Practice

1. What is the equation of a circle centered at  $(-4, 5)$ with a radius of $\sqrt{2}$ ?

2. What are the center and radius of the circle described by the equation:

$x^2-4x+y^2+10y+13=0?$

3. Graph the circle from #2.

1. $h=-4$ , $k=5$ , $r=\sqrt{2}$ . The equation of the circle is $(x-(-4) )^2+(y-5)^2= \left ( \sqrt{2} \right )^2$ . This can be simplified to $(x+4)^2+(y-5)^2=2$ . Notice that the signs of the 4 and the 5 in the equation are opposite from the signs of the 4 and the 5 in the center point.

2. Complete the square to rewrite the equation in graphing form:

$x^2-4x+\underline{4}+y^2+10y+\underline{25} &=-13 \underline{+4+25} \\(x-2)^2+(y+5)^2 &=16$

The center of the circle is  $(2, -5)$ and the radius is 4.

3.

#### Practice

Graph the following circles:

1. $(x+2)^2+(y-5)^2=1$

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

3. $(y+4)^2+(x-3)^2=\frac{1}{4}$

4. $x^2+(y-6)^2=8$

5. $(x-2)^2+y^2=25$

Find the center and radius of the circle described by each equation.

6. $x^2+4x+y^2+4y+7=0$

7. $x^2-2x+y^2+12y+32=0$

8. $x^2-2x+y^2-2y=0$

9. $x^2+8x+y^2+9=0$

10. $x^2+2x+y^2-6y+1=0$

11. Explain why a circle with radius  $r$ that is centered at the origin will have the equation $x^2+y^2=r^2$ .

12. Use the Pythagorean Theorem to help explain why a circle with a center in the second quadrant with radius  $r$ and center  $(h, k)$ will have the equation $(x-h)^2+(y-k)^2=r^2$ . Note that in the second quadrant the value for   $h$ will be negative and the value for   $k$ will be positive.

13. Use the Pythagorean Theorem to help explain why a circle with a center in the third quadrant with radius  $r$ and center  $(h, k)$ will have the equation $(x-h)^2+(y-k)^2=r^2$ . Note that in the third quadrant the value for  $h$ will be negative and the value for   $k$ will be negative.

14. Write the equation of the following circle:

15. Write the equation of the following circle:

### Vocabulary Language: English

center

center

The center of a circle is the point that defines the location of the circle. All points on the circle are equidistant from the center of the circle.
Circle

Circle

A circle is the set of all points at a specific distance from a given point in two dimensions.
Pythagorean Theorem

Pythagorean Theorem

The Pythagorean Theorem is a mathematical relationship between the sides of a right triangle, given by $a^2 + b^2 = c^2$, where $a$ and $b$ are legs of the triangle and $c$ is the hypotenuse of the triangle.

The radius of a circle is the distance from the center of the circle to the edge of the circle.
Vertical Line Test

Vertical Line Test

The vertical line test says that if a vertical line drawn anywhere through the graph of a relation intersects the relation in more than one location, then the relation is not a function.