How to find the equation of a circle

Formula for the equation of a circle

In this lesson we’ll look at the equation for a circle, ${{(x-h)}^{2}}+{{(y-k)}^{2}}={{r}^{2}}$, and how to use it to graph a circle, interpret points on a circle and write an equation given a graph or the special features of a circle.

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A circle can be defined by a center point and a radius of a certain length. In the equation of a circle

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

the center is called $(h,k)$ and the radius is $r$. From the equation, you can see that the circle is the collection of all the $(x,y)$ points that are a distance $r$ from the center, $(h,k)$.

The collection of $(x,y)$ values are the points on the circle’s circumference.

How to find the equation of a circle, given different information about the circle

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Let’s work through four different examples of how to find the equation of a circle

Example

What is the equation of the circle?

We need to find the equation of a circle in the form $(x-h)^2+(y-k)^2=r^2$, which means we need to find the center point and the length of the radius.

The center is at $(2,3)$, so $h=2$ and $k=3$.

Now let’s count from the center to a point on the circumference to find the length of the radius.

The radius is $3$ so $r=3$. Now let’s plug everything into the standard form of a circle.

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

$(x-2)^2+(y-3)^2=9$

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Sometimes we want to know the center and radius of a circle given the equation of the circle.

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Example

What is the center and radius of the circle?

${{x}^{2}}+{{(y-3)}^{2}}=27$

We can rewrite the equation as

${{(x-0)}^{2}}+{{(y-3)}^{2}}=27$

Which lets us identify $h=0$ and $k=3$, so the center is at $(0,3)$. And the radius is then $\sqrt{27}$, so

$r=\sqrt{27}$

$r=\sqrt{9\cdot 3}$

$r=\sqrt{9}\cdot \sqrt{3}$

$r=3\sqrt{3}$

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Sometimes we want to know the -intercepts of a circle.

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A circle can be defined by a center point and a radius of a certain length.

Example

What are the $x$-intercepts of the circle?

${{(x-2)}^{2}}+{{(y+1)}^{2}}=16$

At the $x$-intercepts, $y=0$, so let $y$ be $0$, and rewrite the equation.

${{(x-2)}^{2}}+{{(y+1)}^{2}}=16$

${{(x-2)}^{2}}+{{(0+1)}^{2}}=16$

${{(x-2)}^{2}}+1^2=16$

${{x}^{2}}-4x+4+1=16$

${{x}^{2}}-4x+5=16$

${{x}^{2}}-4x-11=0$

If you can’t factor the equation to solve for $x$ you can use the quadratic formula. In this case, $a=1$, $b=-4$, and $c=-11$.

$x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$

$x=\frac{4\pm \sqrt{{{(-4)}^{2}}-4(1)(-11)}}{2(1)}$

$x=\frac{4\pm \sqrt{60}}{2}=\frac{4\pm \sqrt{4\cdot 15}}{2}=\frac{4\pm 2\sqrt{15}}{2}=2\pm \sqrt{15}$

These are the $x$-intercepts.

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Sometimes to find out information about a circle you’ll need to know how to complete the square.

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Example

Find the center and radius of the circle.

$x^2+y^2+24x+10y+160=0$

In order to find the center and radius, we need to change the equation of the circle into standard form, $(x-h)^2+(y-k)^2=r^2$. In order to get the equation into standard form, we have to complete the square with respect to both variables.

Grouping $x$’s and $y$’s together and moving the constant to the right side, we get

$(x^2+24x)+(y^2+10y)=-160$

Completing the square requires us to take the coefficient on the first degree terms, divide them by $2$, then square the result before adding the result back to both sides.

The coefficient on the $x$ term is $24$, so

$\frac{24}{2}=12\quad\rightarrow\quad12^2=144$

The coefficient on the $y$ term is $10$, so

$\frac{10}{2}=5\quad\rightarrow\quad5^2=25$

So we add $144$ inside the parentheses with the $x$ terms, $25$ inside the parenthesis with the $y$ terms and also add $144$ and $25$ to the right side with the $-160$.

$(x^2+24x +144)+(y^2+10y+25)=-160 + 144+25$

$(x+12)^2+(y+5)^2=9$

Therefore, the center of the circle is at $(h,k)=(-12,-5)$ and its radius is $r=\sqrt{9}=3$. 

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