**Graphing circles** requires two things: the coordinates of the center point, and the radius of a circle. A circle is the set of all points the same distance from a given point, the center of the circle. A radius, $r$, is the distance from that center point to the circle itself.

On a graph, all those points on the circle can be determined and plotted using $(x,y)$ coordinates.

- Graphing a Circle
- Circle Equations
- Using the Center-Radius Form
- How To Graph a Circle Equation
- How To Graph a Circle Using Standard Form

Two expressions show how to plot a circle: the **center-radius form** and the **standard form**. Where $x$ and $y$ are the coordinates for all the circle's points, $h$ and $k$ represent the center point's $x$ and $y$ values, with $r$ as the radius of the circle

The center-radius form looks like this:

${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={r}^{2}$

The standard, or general, form requires a bit more work than the center-radius form to derive and graph. The standard form equation looks like this:

${x}^{2}+{y}^{2}+Dx+Ey+F=0$

In the general form, $D$, $E$, and $F$ are given values, like integers, that are coefficients of the $x$ and $y$ values.

If you are unsure that a suspected formula is the equation needed to graph a circle, you can test it. It must have four attributes:

- The $x$ and $y$ terms must be squared
- All terms in the expression must be positive (which squaring the values in parentheses will accomplish)
- The center point is given as $(h,k)$, the $x$ and $y$ coordinates
- The value for $r$, radius, must be given and must be a positive number (which makes common sense; you cannot have a negative radius measure)

The center-radius form gives away a lot of information to the trained eye. By grouping the $h$ value with the $x{\left(x-h\right)}^{2}$, the form tells you the $x$ coordinate of the circle's center. The same holds for the $k$ value; it must be the $y$ coordinate for the center of your circle.

Once you ferret out the circle's center point coordinates, you can then determine the circle's radius, $r$. In the equation, you may not see ${r}^{2}$, but a number, the square root of which is the actual radius. With luck, the squared $r$ value will be a whole number, but you can still find the square root of decimals using a calculator.

Try these seven equations to see if you can recognize the center-radius form. Which ones are center-radius, and which are just line or curve equations?

- ${\left(x-2\right)}^{2}+{\left(y-3\right)}^{2}=16$
- $5x+3y=6$
- ${\left(x+1\right)}^{2}+{\left(y+1\right)}^{2}=25$
- $y=6x+2$
- ${\left(x+4\right)}^{2}+{\left(y-6\right)}^{2}=49$
- ${\left(x-5\right)}^{2}+{\left(y+9\right)}^{2}=8.1$
- $y={x}^{2}+-6x+3$

**Only equations 1, 3, 5 and 6 are center-radius forms.** The second equation graphs a straight line; the fourth equation is the familiar slope-intercept form; the last equation graphs a parabola.

A circle can be thought of as a graphed line that curves in both its $x$ and $y$ values. This may sound obvious, but consider this equation:

$y={x}^{2}+4$

Here the $x$ value alone is squared, which means we will get a curve, but only a curve going up and down, not closing back on itself. We get a parabolic curve, so it heads off past the top of our grid, its two ends never to meet or be seen again.

Introduce a second $x$-value exponent, and we get more lively curves, but they are, again, not turning back on themselves.

The curves may snake up and down the $y$-axis as the line moves across the $x$-axis, but the graphed line is still not returning on itself like a snake biting its tail.

To get a curve to graph as a circle, you need to change *both* the $x$ exponent *and* the $y$ exponent. As soon as you take the square of both $x$ and $y$ values, you get a circle coming back unto itself!

Often the center-radius form does not include any reference to measurement units like mm, m, inches, feet, or yards. In that case, just use single grid boxes when counting your radius units.

When the center point is the origin $(0,0)$ of the graph, the center-radius form is *greatly* simplified:

${x}^{2}+{y}^{2}={r}^{2}$

For example, a circle with a radius of 7 units and a center at $(0,0)$ looks like this as a formula and a graph:

${x}^{2}+{y}^{2}=49$

If your circle equation is in **standard or general form**, you must first complete the square and then work it into center-radius form. Suppose you have this equation:

${x}^{2}+{y}^{2}-8x+6y-4=0$

Rewrite the equation so that all your $x$-terms are in the first parentheses and $y$-terms are in the second:

$\left({x}^{2}-8x+{?}_{1}\right)+\left({y}^{2}+6y+{?}_{2}\right)=4+{?}_{1}+{?}_{2}$

You have isolated the constant to the right and added the values ${?}_{1}$ and ${?}_{2}$ to both sides. The values ${?}_{1}$ and ${?}_{2}$ are each the number you need in each group to complete the square.

Take the coefficient of $x$ and divide by 2. Square it. That is your new value for ${?}_{1}$:

$\frac{-8}{2}=-4$

${\left(-4\right)}^{2}=16$

${\mathbf{?}}_{\mathbf{1}}\mathbf{=}\mathbf{16}$

Repeat this for the value to be found with the $y$-terms:

$\frac{6}{2}=3$

${3}^{2}=9$

${\mathbf{?}}_{\mathbf{2}}\mathbf{=}\mathbf{9}$

Replace the unknown values ${?}_{1}$ and ${?}_{2}$ in the equation with the newly calculated values:

$\left({x}^{2}-8x+{\mathbf{16}}\right)+\left({y}^{2}+6y+{\mathbf{9}}\right)=4+{\mathbf{16}}+{\mathbf{9}}$

**Simplify:**

$\left({x}^{2}-8x+16\right)+\left({y}^{2}+6y+9\right)=29$

${\left(x-4\right)}^{2}+{\left(y+3\right)}^{2}=29$

You now have the center-radius form for the graph. You can plug the values in to find this circle with center point $\left(-4,3\right)$ and a radius of $5.385$ units (the square root of 29):

In practical terms, remember that the center point, while needed, is not actually part of the circle. So, when actually graphing your circle, mark your center point very lightly. Place the easily counted values along the $x$ and $y$ axes, by simply counting the radius length along the horizontal and vertical lines.

If precision is not vital, you can sketch in the rest of the circle. If precision matters, use a ruler to make additional marks, or a drawing compass to swing the complete circle.

You also want to mind your negatives. Keep careful track of your negative values, remembering that, ultimately, the expressions must all be positive (because your $x$-values and $y$-values are squared).

Instructor: **Malcolm M.**

Malcolm has a Master's Degree in education and holds four teaching certificates. He has been a public school teacher for 27 years, including 15 years as a mathematics teacher.

Malcolm has a Master's Degree in education and holds four teaching certificates. He has been a public school teacher for 27 years, including 15 years as a mathematics teacher.

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