**Equation of Circles**

Let’s review what we already know about circles.

**Definition:** A circle is a locus (set) of points in a plane equidistant from a fixed point.

Now, if we “multiply out” the above example (x-2)^{2 }+ (y+5)^{2 }= 9 we will get:

(x-2)^{2 }+ (y+5)^{2 }= 9

(x^{2}-4x+4) +( y^{2}+10y+25) = 9

x^{2}-4x+4+ y^{2}+10y+25 = 9

x^{2}+ y^{2}-4x+10y+20= 0

x^{2}+y^{2}+Cx+Dy+E = 0

When multiplied out, we obtain the “general form” of the equation of a circle. Notice that in this form we can clearly see that the equation of a circle has both x^{2} and y^{2} terms and these terms have the same coefficient (usually 1).

When the equation of a circle appears in “general form”, it is often beneficial to convert the equation to “center-radius” form to easily read the center coordinates and the radius for graphing.

**Examples:**

**1. Convert x ^{2}+ y^{2}-4x-6y+8= 0 into center-radius form.**

This conversion requires use of the technique of completing the square.

We will be creating two perfect square trinomials within the equation.

- Start by grouping the x related terms together and the y related terms together. Move any numerical constants (plain numbers) to the other side.
- Get ready to insert the needed values for creating the perfect square trinomials. Remember to balance both sides of the equation.
- Find each missing value by taking half of the “middle term” and squaring. This value will always be positive as a result of the squaring process.
- Rewrite in factored form.

You can now read that the center of the circle is at (2, 3) and the radius is /5

**2. How do the coordinates of the center of a circle relate to C and D when the equation of the circle is in the general form x ^{2}+ y^{2}+Cx+Dy+E = 0.**

Let’s make some observations. Re-examine our previous equations in general form and center-radius form. Do you see a relationship between the center coordinates and C and D?

**3. Write the equation of a circle whose diameter has endpoints (4, -1) and (-6, 7).**

**4. Write the equation for the circle shown below if it is shifted 3 units to the right and 4 units up.**

A shift of 3 units to the right and 4 units up places the center at the point (3, 4). The radius of the circle can be seen from the graph to be 5.

Equation:(x-3)^{2 }+ (y-4)^{2 }= 25

**5. Convert 2x ^{2}+ 2y^{2}+6x-8y+12= 0 into center-radius form.**

Whoa!!! This equation looks different. Are we sure this is a circle???

In this equation, both the x and y terms appear in squared form and their coefficients (the numbers in front of them) are the same. Yes, we have a circle here! We will, however, have to deal with the coefficients of 2 before we can complete the square.

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