How to find the center and radius of a circle from its equation

What is the standard form for the equation of a circle?

In this lesson we’ll look at how to write the equation of a circle in standard form in order to find the center and radius of the circle.

Hi! I'm krista.

I create online courses to help you rock your math class. Read more.

The standard form for the equation of a circle is

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

where $r$ is the radius and $(h,k)$ is the center.

Sometimes in order to write the equation of a circle in standard form, you’ll need to complete the square twice, once for $x$ and once for $y$.

Putting the equation of a circle into standard form, and then finding its center and radius from the standard form

---

Take the course

Want to learn more about Algebra 2? I have a step-by-step course for that. :)

---

Step-by-step examples of finding the center and radius of circles

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, $(x-h)^2+(y-k)^2=r^2$, where $h$ and $k$ are the coordinates of the center and $r$ is the radius.

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 $2$, and then square the result before adding the result back to both sides.

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

$\frac{24}{2}=12$

$12^2=144$

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

$\frac{10}{2}=5$

$5^2=25$

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

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

Factor the parentheses and simplify the right side.

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

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

---

Let’s do another.

---

Sometimes in order to write the equation of a circle in standard form, you’ll need to complete the square twice, once for x and once for y.

Example

What is the center and radius of the circle?

$6x^2+6y^2+12x-13=0$

In order to find the center and radius, we need to change the equation of the circle into standard form, $(x-h)^2+(y-k)^2=r^2$, where $h$ and $k$ are the coordinates of the center and $r$ is the radius.

In order to get the equation into standard form, we have to complete the square with respect to $x$. The $y$ term is already a perfect square.

Let’s begin by collecting like terms and moving the $-13$ to the right.

$6x^2+12x+6y^2=13$

Our next step is to remove the coefficients of the second degree terms of the $x$ variable and $y$ variable. First, we’ll factor out a $6$ then divide by $6$ on both sides.

$6(x^2+2x+y^2)=13$

$x^2+2x+y^2=\frac{13}{6}$

Now complete the square of the $x$ terms. The $y$ term is already a perfect square.

$(x^2+2x)+y^2=\frac{13}{6}$

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

The coefficient on $x$ is $2$, so

$\frac{2}{2}=1$

$1^2=1$

We’ll therefore add $1$ to both sides, and get

$(x^2+2x+1)+y^2=\frac{13}{6}+1$

Factor the $x$ terms and simplify the right hand side.

$(x+1)^2+y^2=\frac{19}{6}$

If you want, you may also write the equation as

$(x+1)^2+(y+0)^2=\frac{19}{6}$

The center of the circle $(h,k)$ is $(-1,0)$ and the radius is $\sqrt{19/6}$. Rule out $-\sqrt{19/6}$ because a radius can't be negative.

---

Get access to the complete Algebra 2 course

---