Read and write circle equations

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Lesson progressPractice problems 0/4

Difficulty: Beginner
Estimated time: 27 minutes
Domains: Geometry and Trigonometry
Techniques: Circle-equationsStandard-formCenter-and-radiusPoint-classificationTranslations

What you’ll learn

Read the center and radius straight from an equation like $(x+4)^2+(y-3)^2=49$ .
Write a circle’s equation from its center and radius, or from its center and one point on it.
Tell whether a point is inside, on, or outside a circle with one quick calculation.
Predict how a circle moves when you change the numbers in its equation.
Spot when a line crossing a circle means you should graph both and click the intersections.

Prerequisites

Find perimeter and area of plane figures0%
Recall that a radius runs from the center to any point on a circle.

Why this matters on the SAT

Turn a center and radius into one equation

Every circle equation holds two facts: where the center is and how far the circle reaches. On the SAT, you’ll read those facts, build an equation from them, or predict how the graph moves when they change. Here’s a typical question.

Solution to the example

Start with the radius. To get from the center $(2,-1)$ to the point $(5,3)$ , you go $5-2=3$ units right and $3-(-1)=4$ units up. Those two moves are the legs of a right triangle, and the radius is its long side, so

r^2=3^2+4^2=25.

That makes $r=5$ , which matches the figure.

Now place the center. With $(h,k)=(2,-1)$ , the standard form

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

becomes

(x-2)^2+(y-(-1))^2=25,

which simplifies to

(x-2)^2+(y+1)^2=25.

The answer is A.

Each wrong choice is a classic slip. B flips the signs of the center, C forgets to square the radius, and D puts the center at the point on the circle. By the end of this lesson, you’ll see each one coming.

SAT example

The horizontal and vertical changes from the center to the marked point are 3 and 4.

The graph shows a circle with center $(2,-1)$ . The point $(5,3)$ lies on the circle. Which equation represents the circle?

A
$(x-2)^2+(y+1)^2=25$
B
$(x+2)^2+(y-1)^2=25$
C
$(x-2)^2+(y+1)^2=5$
D
$(x-5)^2+(y-3)^2=25$

Standard form measures distance

A circle is every point that sits the same distance from one center point. The standard form of a circle writes that idea as an equation:

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

Here $(h,k)$ is the center and $r$ is the radius, which is always positive.

To see why it works, go back to the SAT example. Its circle is $(x-2)^2+(y+1)^2=25$ , and $(5,3)$ is on it. Plug that point into the left side:

(5-2)^2+(3+1)^2=3^2+4^2=25.

The $3$ and the $4$ are how far the point sits right of the center and above it. Square them and add, and you get the point’s distance from the center, squared. That’s the Pythagorean theorem on a grid. Every point on this circle gives exactly $25$ , because every point on it is $5$ units from the center.

The same thing happens on any circle. $x-h$ is how far a point sits right or left of the center, and $y-k$ is how far up or down. So standard form says that every point on the circle is exactly $r$ units from $(h,k)$ .

Now let’s read one:

(x+4)^2+(y-3)^2=49.

Standard form has minus signs built in, so read $x+4$ as $x-(-4)$ , and read $49$ as $7^2$ :

(x-(-4))^2+(y-3)^2=7^2.

The center is $(-4,3)$ and the radius is $7$ .

Common mistake:

The signs you see are the opposite of the center’s coordinates. Copy them and you’d get $(4,-3)$ , which is wrong. To get it right every time, ask what makes each parenthesis zero: $x+4=0$ when $x=-4$ , and $y-3=0$ when $y=3$ . So the center is $(-4,3)$ .

Check your understanding:

What are the center and radius of $(x-6)^2+(y+2)^2=81$ ?

Write the equation from what you know

When you know the center and the radius, put them straight into standard form:

\text{center }(h,k),\ \text{radius }r \quad\longrightarrow\quad (x-h)^2+(y-k)^2=r^2.

For a center at $(-3,5)$ and radius $4$ :

\begin{aligned} (x-h)^2+(y-k)^2&=r^2\\[1.4em] (x-(-3))^2+(y-5)^2&=4^2\\[1.4em] (x+3)^2+(y-5)^2&=16. \end{aligned}

Sometimes you get the center and one point on the circle instead. The radius is the distance from the center to that point, so use the same right-triangle idea:

r^2=(\text{horizontal change})^2+(\text{vertical change})^2.

That hands you $r^2$ directly, and $r^2$ is exactly what the equation needs. You don’t need its square root to write the equation.

Common mistake:

When a question gives you the radius, it’s easy to put it straight on the right side. Square it first: a radius of $4$ puts $4^2=16$ there, not $4$ .

Example: Write from a center and point

Worked example

The distance from the center to

(2,0)

is the radius.

A circle in the $xy$ -plane has center $(-2,3)$ and passes through the point $(2,0)$ . Which equation represents the circle?

A
$(x-2)^2+(y+3)^2=25$
B
$(x+2)^2+(y-3)^2=25$
C
$(x+2)^2+(y-3)^2=5$
D
$(x+2)^2+(y-3)^2=7$

Step 1

Place the center

The center $(-2,3)$ gives $h=-2$ and $k=3$ . Put them into the left side of standard form:

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

Step 2

Find the squared radius

To get from $(-2,3)$ to $(2,0)$ , you go $4$ units right and $3$ units down. So the horizontal change is $4$ and the vertical change is $-3$ :

\begin{aligned} r^2&=4^2+(-3)^2\\[1.4em] &=16+9\\[1.4em] &=25. \end{aligned}

The minus sign on the $-3$ doesn’t matter, because squaring makes it positive.

Step 3

Complete and check the equation

The equation is

(x+2)^2+(y-3)^2=25.

Check it with the point you were given, $(2,0)$ :

(2+2)^2+(0-3)^2=4^2+(-3)^2=25.

The point fits, so the answer is B. Choice C stops at $r=5$ without squaring it, and choice D adds $4+3$ without squaring either change.

Check your understanding:

Keep the center at $(-2,3)$ , but make the point on the circle $(-2,10)$ . What is the new equation?

Inside, on, or outside? Make one comparison

Is the point $(5,2)$ inside, on, or outside the same circle,

(x-2)^2+(y+1)^2=25?

Plug the point into the left side, just as you did with $(5,3)$ :

(5-2)^2+(2+1)^2=3^2+3^2=18.

Points on the circle give exactly $25$ . This one gives only $18$ , so it’s closer to the center than the circle is. It’s inside.

Compare with $r^2=25$ , not with $r=5$ . Next to $5$ , the $18$ would look too big and wrongly put the point outside.

The same test works for any point. Plug it into the left side and compare the result with $r^2$ :

less than $r^2$ : the point is inside the circle;
equal to $r^2$ : the point is on the circle;
greater than $r^2$ : the point is outside the circle.

Common mistake:

The circle is only the edge: think of a ring, not a filled-in disk. The equation, with its equals sign, describes that edge and nothing inside it. So a point is on the circle only when it gives exactly $r^2$ . The point $(5,2)$ is inside the circle, not on it.

Check your understanding:

Is $(6,2)$ inside, on, or outside $(x-2)^2+(y+1)^2=25$ ?

Move the center, move the graph

Now slide that same circle,

(x-2)^2+(y+1)^2=25,

$3$ units right and $4$ units up. Its center starts at $(2,-1)$ . Move the center, and the whole circle comes with it:

(2+3,-1+4)=(5,3).

Put the new center into standard form:

(x-5)^2+(y-3)^2=25.

The right side stays $25$ because the radius is still $5$ . That’s choice D from the SAT example: the right size, in the wrong place.

So changing $h$ or $k$ slides the circle:

increase $h$ to move it right, and decrease $h$ to move it left;
increase $k$ to move it up, and decrease $k$ to move it down.

Here’s the tricky part. Moving up $4$ turned $y+1$ into $y-3$ , so the number you see went down while the circle went up. That’s the sign flip again, and it’s why you move the center, not the numbers you see.

You can do all of this by hand, and that’s the fastest way. Desmos is still a good check, because it shows both circles at once, one slid over from the other.

Try it yourself:

Before you type $(x-5)^2+(y-3)^2=25$ , say where its center should be and how big it should be. Then add it on a new line, keeping the original circle, and compare the two.

Desmos takes the lead when a line or another condition meets a circle at points that aren’t obvious: graph both equations first and click every intersection. Solve linear-nonlinear systems covers those questions, including when exact answers still need algebra. For more Desmos practice with circles, try Circle equations in the graph.

Calculator loads as you approach

The starting circle: center (2, -1), radius 5. Add the new circle on its own line.

Practice problems

Each of these takes a line or two by hand. If circle questions have tripped you up before, take them slowly: every one uses a move you’ve just practiced.

Write from a center and radius

Practice problem

A circle in the $xy$ -plane has center $(4,-3)$ and radius $6$ . Which equation represents the circle?

Calculator loads as you approach

Graph your choice if you want to check its center and radius.

Read the center and radius

Practice problem

Which choice gives the center and radius of the circle

(x+7)^2+(y-2)^2=121?

Calculator loads as you approach

Read the center and radius first. Graph only if you want a check.

Classify a point

Practice problem

Compare the point’s squared distance from the center with

r^2=25

Relative to the circle

(x-1)^2+(y+2)^2=25,

where is the point $(6,2)$ ?

Calculator loads as you approach

A graph can show you roughly where the point is. Put the point into the equation to be sure.

Translate the equation

Practice problem

The circle

(x+1)^2+(y-4)^2=9

is translated $5$ units to the right and $2$ units down. Which equation represents the translated circle?

Calculator loads as you approach

Move the center first. Graph both circles if you want to compare them.

Finish the lesson

4 practice examples left

Finish the remaining questions correctly to complete this lesson.

Quick recap

Standard form is $(x-h)^2+(y-k)^2=r^2$ : every point on the circle is $r$ units from the center $(h,k)$ .
The signs inside the parentheses are opposite the center’s coordinates. Ask what makes each parenthesis zero.
The right side is $r^2$ , so the radius is its positive square root.
From a center and a point, square and add the horizontal and vertical changes to get $r^2$ .
To tell where a point is, plug it into the left side and compare with $r^2$ , not $r$ : less is inside, equal is on, and greater is outside.
To move a circle, move its center. The right side stays the same because the size doesn’t change.
Work standard form by hand, and use Desmos to check a move. When a line meets a circle at points that aren’t obvious, graph both and click every intersection.

Next lesson

Complete the square for a circle

Rewrite an expanded circle equation so its center and radius become visible.

Start next lesson

Practice

Practice this lesson

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

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