Complete the square for a circle

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Lesson progressPractice problems 0/3

Difficulty: Intermediate
Estimated time: 28 minutes
Domains: Geometry and Trigonometry
Techniques: Completing-the-squareGraph-expanded-circleDivide-shared-coefficient

What you’ll learn

Spot an expanded circle equation, one with $x^2$ , an $x$ -term, $y^2$ , and a $y$ -term but no squared parentheses.
Graph the whole equation in Desmos and find an integer center, radius, or diameter from the highest and lowest points it marks.
Group the $x$ -terms and the $y$ -terms.
Complete both squares and keep the equation balanced.
Divide out a coefficient that $x^2$ and $y^2$ share before you complete the squares.
Write exact standard form, and an exact radical radius, when the question asks for them.

Prerequisites

Why this matters on the SAT

Reveal the circle hidden in the algebra

Multiply out a circle equation and the center and radius disappear from view. You can get them back two ways. Completing the square by hand puts the equation back into standard form. Graphing it in Desmos shows you the circle itself. What the question asks for decides which way to go. This one asks for an equation, so the algebra is the way in.

Solution to the example

Move the $-12$ to the right side, and group the $x$ -terms and the $y$ -terms:

(x^2+6x)+(y^2-4y)=12.

Now complete each square. Half of $6$ is $3$ , so add $3^2=9$ to the $x$ -group. Half of $-4$ is $-2$ , so add $(-2)^2=4$ to the $y$ -group. Add both numbers to the right side too:

\begin{aligned} (x^2+6x+9)+(y^2-4y+4)&=12+9+4,\\[1.4em] (x+3)^2+(y-2)^2&=25. \end{aligned}

The answer is B.

Each wrong choice is a common slip. A flips the signs inside the parentheses, C adds $9$ and $4$ on the left but not on the right, and D uses $6$ and $-4$ instead of their halves.

Since this one is multiple-choice, you could also graph the given equation and each promising choice in Desmos, then keep the choice whose circle lands exactly on top of the original.

SAT example

Which equation is equivalent to

x^2+y^2+6x-4y-12=0?

A
$(x-3)^2+(y+2)^2=25$
B
$(x+3)^2+(y-2)^2=25$
C
$(x+3)^2+(y-2)^2=12$
D
$(x+6)^2+(y-4)^2=25$

Complete two squares, not one

Let’s look at that move on its own. Take $x^2-8x$ . Half of $-8$ is $-4$ , and $(-4)^2=16$ , so add $16$ :

x^2-8x+16=(x-4)^2.

Now $y^2+6y$ . Half of $6$ is $3$ , and $3^2=9$ , so add $9$ :

y^2+6y+9=(y+3)^2.

Half it, square it, add it. The half goes inside the parentheses, and its square is the number you add. In general, when the squared term has coefficient $1$ , $u^2+bu$ works like this, where $u$ is $x$ or $y$ :

u^2+bu+\left(\frac b2\right)^2 = \left(u+\frac b2\right)^2.

A circle equation needs this twice. The $x$ -term’s coefficient builds the $x$ -square, and the $y$ -term’s coefficient builds the $y$ -square.

Check your understanding:

What do you add to complete $x^2+10x$ and $y^2-12y$ , and what squares do you get?

Keep the equation balanced

An equation is a balance. Add a number to one side only, and the two sides stop being equal, so you’re no longer describing the same circle. That’s why every number you add on the left also goes on the right.

Start with

x^2+y^2-8x+6y=11.

Group the terms and leave a space in each group for the number you’ll add:

(x^2-8x+\square)+(y^2+6y+\square)=11.

You found those numbers above:

\left(\frac{-8}{2}\right)^2=16 \qquad\text{and}\qquad \left(\frac{6}{2}\right)^2=9.

Add both to both sides:

\begin{aligned} (x^2-8x+16)+(y^2+6y+9)&=11+16+9,\\[1.4em] (x-4)^2+(y+3)^2&=36. \end{aligned}

Now standard form shows the center, $(4,-3)$ , and the radius, $6$ .

Common mistake:

It’s easy to complete both squares on the left and leave the right side alone. Here that gives $(x-4)^2+(y+3)^2=11$ : the right center, but the wrong radius. Write the two added numbers on the right before you factor, so both additions sit in one place you can check.

Try it yourself:

Before reading on, expand $(x-4)^2+(y+3)^2=36$ and move the constants to the right. Do you get back $x^2+y^2-8x+6y=11$ ?

Expanding runs the steps backward, which makes it a reliable check:

\begin{aligned} (x-4)^2+(y+3)^2&=36,\\[1.4em] x^2-8x+16+y^2+6y+9&=36,\\[1.4em] x^2+y^2-8x+6y&=11. \end{aligned}

Example: Find the center and radius

Worked example

The equation

x^2+y^2-10x+8y-8=0

represents a circle in the $xy$ -plane. Which choice gives the circle’s center and radius?

A
Center $(-5,4)$ and radius $7$
B
Center $(5,-4)$ and radius $49$
C
Center $(5,4)$ and radius $7$
D
Center $(5,-4)$ and radius $7$

Step 1

Graph the whole equation

This time the question wants two numbers, not an equation. A graph can show both, so start in Desmos. Type the whole equation exactly as it’s written. It already has both $x$ and $y$ in it, so Desmos can graph it as it stands.

Step 2

Find the highest and lowest points

Select the circle. The testing calculator marks its highest and lowest points:

(5,3) \qquad\text{and}\qquad (5,-11).

The top and bottom of a circle sit straight above and below its center, so these two points are the ends of a vertical diameter.

Step 3

Use the vertical diameter

The center is halfway between the two points. Both have $x$ -coordinate $5$ , and halfway between their $y$ -coordinates is

\frac{3+(-11)}{2}=-4.

So the center is $(5,-4)$ . The radius is half the diameter, so it’s half the vertical distance:

\frac{3-(-11)}{2}=7.

Step 4

Match the answer and spot the traps

The center is $(5,-4)$ and the radius is $7$ , so the answer is D.

Choice B gives $49$ , which is $r^2$ , not the radius. Choice C misses the sign flip: $y+4$ is $y-(-4)$ , so the center’s $y$ -coordinate is $-4$ .

If the question had asked for the equation itself, you’d complete the square to get

(x-5)^2+(y+4)^2=49.

Check your understanding:

Suppose a selected circle’s marked highest and lowest points are $(2,9)$ and $(2,-3)$ . What are its center and radius? And why use the marked points instead of reading them off the grid?

Common mistake:

Putting y= in front of the expanded equation turns it into a different relation, so you’d graph the wrong thing. Type the equation exactly as given, select the circle, and work from its marked points.

Common mistake:

When you complete the square by hand for an exact answer, check two details. The center’s signs are the opposite of the ones in the parentheses, and the right side is $r^2$ , not $r$ . Match each parenthesis to $x-h$ or $y-k$ , then take the positive square root of the right side.

Calculator loads as you approach

Select the circle to see its highest and lowest points.

Divide a shared coefficient first

Half it, square it, add it assumes that $x^2$ and $y^2$ each have coefficient $1$ . If they share a different coefficient, divide the entire equation by it first.

Take

4x^2+4y^2+24x-32y=20.

Divide every term by $4$ :

x^2+y^2+6x-8y=5.

Now complete the two squares:

\begin{aligned} (x^2+6x+9)+(y^2-8y+16)&=5+9+16,\\[1.4em] (x+3)^2+(y-4)^2&=30. \end{aligned}

The center is $(-3,4)$ , and the radius is

r=\sqrt{30}.

That’s already in simplest form, because $30$ has no perfect-square factor greater than $1$ . A graph would only show you decimals here, so an exact radical like this one needs the algebra.

Common mistake:

It’s tempting to divide only $4x^2$ and $4y^2$ . But you’re dividing the whole equation, so every term gets divided by $4$ , including $24x$ , $-32y$ , and the $20$ on the right.

Check your understanding:

What’s your first step with $3x^2+3y^2-12x+18y=6$ , and what equation does it give you?

Let the question pick the method

Before you start, look at what the question asks for.

Graph it in Desmos

You need an integer center, radius, or diameter, as in the worked example.
You’re asked which graph shows the equation.
You need another number that the highest and lowest points give you, such as a diameter to scale up.

Complete the square by hand

You need an equivalent equation in standard form, as in the SAT example.
The radius has to be an exact radical like $\sqrt{30}$ , which the graph won’t write for you.
The squared terms share a coefficient, as in $4x^2+4y^2$ , and you need exact form once you divide it out.

If both ways would work, pick the one with less setup and less room to misread.

If the graph already shows the center, radius, or picture you need, skip the algebra. Completing the square and then graphing only as a check doubles the work.

For more Desmos practice with circles, try Circle equations in the graph.

Practice problems

Before each one, ask what it wants: numbers a graph can show, or an exact equation. Each calculator starts blank and keeps your work.

Rewrite in standard form

Practice problem

Which equation is equivalent to

x^2+y^2+4x-12y=9?

Calculator loads as you approach

Complete the squares, or graph the original and a choice to see if their circles match.

Find an integer radius

Practice problem

The equation

x^2+y^2+10x-4y+13=0

represents a circle in the $xy$ -plane. What is the radius of the circle?

Your answer

Calculator loads as you approach

Graph the whole equation and use the circle’s highest and lowest points.

Transfer the center and scale

Practice problem

Circle $A$ is represented by the equation

x^2+y^2+4x-10y+4=0.

Circle $B$ has the same center as circle $A$ , and the diameter of circle $B$ is $3$ times the diameter of circle $A$ . The point $(a,-4)$ lies on circle $B$ , where $a>0$ . What is the value of $a$ ?

Your answer

Calculator loads as you approach

Graph circle A and select it. Its highest and lowest points give you the diameter to scale.

Finish the lesson

3 practice examples left

Finish the remaining questions correctly to complete this lesson.

Quick recap

An expanded circle equation has $x^2$ , an $x$ -term, $y^2$ , and a $y$ -term, with no squared parentheses.
If the question wants an integer center, radius, or diameter, graph the whole equation first.
Graph first, too, when it asks which graph matches, or needs another number the marked points give you.
Select the circle and treat its marked highest and lowest points as the ends of a vertical diameter. Don’t estimate from the grid.
If the question wants exact standard form or an exact radical, complete the square.
Move the constant, group by variable, and complete the $x$ -square and the $y$ -square separately.
Half it, square it, add it: adding $\left(\frac b2\right)^2$ to $u^2+bu$ makes $\left(u+\frac b2\right)^2$ .
Add every completing number to the right side too, so the equation stays balanced.
If $x^2$ and $y^2$ share a coefficient other than $1$ , divide the whole equation by it first.
Match the result to $(x-h)^2+(y-k)^2=r^2$ .
The center’s signs are the opposite of the ones in the parentheses. The radius is the positive square root of the right side.

Next lesson

Build circles from coordinate information

Use midpoint, distance, diameter endpoints, and other coordinate conditions to construct a circle.

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Practice

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77 SAT questions use what this lesson teaches. Practice a few in a study session at the difficulty you choose.

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