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## Integrated math 2

### Course: Integrated math 2>Unit 12

Lesson 4: Expanded equation of a circle

# Circle equation review

Review the standard and expanded forms of circle equations, and solve problems concerning them.

## What is the standard equation of a circle?

$\left(x-h{\right)}^{2}+\left(y-k{\right)}^{2}={r}^{2}$
This is the general standard equation for the circle centered at $\left(h,k\right)$ with radius $r$.
Circles can also be given in expanded form, which is simply the result of expanding the binomial squares in the standard form and combining like terms.
For example, the equation of the circle centered at $\left(1,2\right)$ with radius $3$ is $\left(x-1{\right)}^{2}+\left(y-2{\right)}^{2}={3}^{2}$. This is its expanded equation:
$\begin{array}{rl}\left(x-1{\right)}^{2}+\left(y-2{\right)}^{2}& ={3}^{2}\\ \\ \left({x}^{2}-2x+1\right)+\left({y}^{2}-4y+4\right)& =9\\ \\ {x}^{2}+{y}^{2}-2x-4y-4& =0\end{array}$

## Practice set 1: Using the standard equation of circles

Problem 1.1
$\left(x+4{\right)}^{2}+\left(y-6{\right)}^{2}=48$
What is the center of the circle?
$\left($
$,$
$\right)$
units

Want to try more problems like this? Check out this exercise and this exercise.

## Practice set 2: Writing circle equations

Problem 2.1
A circle has a radius of $\sqrt{13}$ units and is centered at $\left(-9.3,4.1\right)$.
Write the equation of this circle.

Want to try more problems like this? Check out this exercise.

## Practice set 3: Using the expanded equation of circles

To interpret the expanded equation of a circle, we should rewrite it in standard form using the method of "completing the square."
Consider, for example, the process of rewriting the expanded equation ${x}^{2}+{y}^{2}+18x+14y+105=0$ in standard form:
$\begin{array}{rl}{x}^{2}+{y}^{2}+18x+14y+105& =0\\ \\ {x}^{2}+{y}^{2}+18x+14y& =-105\\ \\ \left({x}^{2}+18x\right)+\left({y}^{2}+14y\right)& =-105\\ \\ \left({x}^{2}+18x+81\right)+\left({y}^{2}+14y+49\right)& =-105+81+49\\ \\ \left(x+9{\right)}^{2}+\left(y+7{\right)}^{2}& =25\\ \\ \left(x-\left(-9\right){\right)}^{2}+\left(y-\left(-7\right){\right)}^{2}& ={5}^{2}\end{array}$
Now we can tell that the center of the circle is $\left(-9,-7\right)$ and the radius is $5$.
Problem 3.1
${x}^{2}+{y}^{2}-10x-16y+53=0$
What is the center of this circle?
$\left($
$,$
$\right)$
What is the radius of this circle?
units

Want to try more problems like this? Check out this exercise and this exercise.