Circle equations in the graph

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Type (x - h)^2 + (y - k)^2 = r^2 and Desmos draws the circle. Read the center and radius, test points, and find where a line crosses or touches it.

Difficulty: Intermediate
Estimated time: 55 minutes
Domains: Geometry and Trigonometry

What you’ll learn

Graph a whole circle equation without solving it for $y$ .
Read the center and radius from center-radius form.
Use the center and radius to find how far the circle reaches left, right, up, and down.
Tell whether a point is inside, on, or outside a circle.
Turn a condition about a point into a second graph.
Read where a line meets a circle, and spot when the line is tangent.
Decide when a graph helps and when working by hand is faster.

Prerequisites

Why this matters on the SAT

Turn a circle equation into a picture

On the SAT, a circle can show up as a long, messy equation, and you can't see where its center is just by looking. Type the whole equation into Desmos, though, and it draws the circle for you. You can click its highest and lowest points and see exactly where it meets a line.

SAT example

In the $xy$ -plane, the graph of

x^2+y^2+8x-6y-75=0

is a circle. The horizontal line $y=9$ intersects the circle at two points. What is the positive $x$ -coordinate of an intersection point?

Fast Desmos solution

Type the circle equation on one line and y=9 on another. Click the two points where they meet. Desmos shows

(-12,9) \qquad\text{and}\qquad (4,9).

The question wants the positive $x$ -coordinate, which is $4$ , so the answer is C.

Notice who did what. Desmos found both points, but it has no idea which one you need. The question asked for the positive $x$ -coordinate, so you're the one who rules out $-12$ . Keep that split in mind whenever you graph: Desmos finds the points, and the question picks the answer.

Calculator loads as you approach

Click both points where the graphs meet, then go back to the question to pick the positive x-coordinate.

When should you use Desmos?

A graph helps most when the equation is expanded or the question adds a condition you can draw. When the algebra is already clean, working by hand is usually faster.

Graph the circle when…

The equation is expanded, like $x^2+y^2+8x-6y-75=0$ , so you can’t see the center at a glance.
The question asks for one coordinate of a point on the circle, like “the positive $x$ -coordinate.”
You need to test whether points lie on the circle, such as four answer choices that are all points.
A line or a condition like $x=5$ crosses the circle, and the answer sits where they meet.
The answer depends on whether a line crosses the circle, touches it once, or misses it.

Work it out by hand first when…

The equation is already in center-radius form, like $(x+4)^2+(y-3)^2=49$ .
The question asks only for the center, the radius, the diameter, or how far the circle reaches in some direction.
One quick substitution tells you exactly whether a point is on the circle.
Desmos shows a decimal like $5.646$ , but the answer has to be exact, like $3+\sqrt7$ .

Plenty of questions need both: a graph to find the points, then a line or two of algebra to match an exact answer.

Circles given by the endpoints of a diameter come later, in circle geometry with distance and midpoint, and circles through three given points come in geometry to coordinates to circle regression.

Check your understanding:

Which one would you start in Desmos: finding the radius of $(x-2)^2+(y+5)^2=36$ , or finding where $x^2+y^2-6x+4y-12=0$ meets $y=2x-1$ ?

Enter the whole circle equation

Type it exactly as written

You've probably graphed lots of equations that start with y=. A circle doesn't fit that pattern. Draw a vertical line through most $x$ -values across its width, and it hits the circle twice, once above the center and once below. So a circle usually isn't one function of $x$ .

That's fine, because Desmos can graph the whole equation as it is. Type it in exactly as written, with no y= in front:

x^2 + y^2 - 4x + 2y - 20 = 0

Now select the circle. The current College Board calculator marks its highest and lowest points:

(2,4) \qquad\text{and}\qquad (2,-6).

The highest and lowest points sit straight above and below the center, so they're the ends of a vertical diameter. The center is halfway between them, at $(2,-1)$ . The diameter is $4-(-6)=10$ units long, so the radius is $5$ .

Here's a catch: Desmos doesn't reliably mark the leftmost and rightmost points, so you can't click them the same way. Don't guess them from the grid. When the equation is in center-radius form, you can work them out exactly from the center and radius instead.

Try it yourself:

Swap in x^2+y^2+2x-8y-8=0. Before you click, guess the $y$ -coordinates of the highest and lowest points. Then select the circle and see how close you were.

Check your understanding:

For $x^2+y^2+2x-8y-8=0$ , what are the highest and lowest $y$ -coordinates on the graph?

Common mistake:

Typing y=x^2+y^2-4x+2y-20=0. It’s an easy habit to fall into, since so many graphs start with y=. But this equation already has both $x$ and $y$ in it. The extra y= turns it into a different relation, so Desmos no longer draws the circle you were given.

Calculator loads as you approach

Select the circle to see its highest and lowest points.

Read center-radius form before graphing

Watch the signs

Center-radius form looks like this:

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

The center is $(h,k)$ , and $r$ is the radius, a positive number.

The tricky part is the signs. The form has minus signs built in, so a plus sign in the equation means a negative coordinate. Take

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

Rewrite $x+4$ as $x-(-4)$ , and $49$ as $7^2$ :

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

So the center is $(-4,3)$ , not $(4,-3)$ , and the radius is $7$ .

Use the radius to find the edges

A circle reaches exactly $r$ units from its center in every direction, including straight left, right, up, and down. So for a circle with center $(h,k)$ and radius $r$ :

Edge	Coordinate
least $x$ (left edge)	$h-r$
greatest $x$ (right edge)	$h+r$
least $y$ (bottom)	$k-r$
greatest $y$ (top)	$k+r$

For the circle above:

\begin{aligned} \text{least }x&=-4-7=-11,\\[1.4em] \text{greatest }x&=-4+7=3,\\[1.4em] \text{least }y&=3-7=-4,\\[1.4em] \text{greatest }y&=3+7=10. \end{aligned}

Four quick subtractions and additions, and you have exact values. That's faster and more reliable than hunting for the leftmost and rightmost points on the graph.

Try it yourself:

Change the right side from $49$ to $81$ . Work out the new radius and all four edges first, then look at the new circle to check.

Check your understanding:

What are the center and radius of $(x-11)^2+(y+8)^2=196$ ?

Common mistake:

Calling $49$ the radius of $(x+4)^2+(y-3)^2=49$ . The right side is $r^2$ , the radius squared. So the radius is $\sqrt{49}=7$ .

Calculator loads as you approach

The marked center is (-4, 3). Every point on the circle is 7 units from it.

Test a point against the circle

Plug the point in and compare

Say you want to know where a point $(a,b)$ sits compared with the circle

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

Plug the point into the left side:

(a-h)^2+(b-k)^2.

That value is the point's squared distance from the center. So compare it with $r^2$ , the squared radius:

What you get	Where the point is
less than $r^2$	inside the circle
equal to $r^2$	on the circle
greater than $r^2$	outside the circle

Let's try it with

(x-1)^2+(y+3)^2=25.

For $(4,1)$ :

(4-1)^2+(1+3)^2=3^2+4^2=25,

so the point is on the circle.

For $(5,1)$ :

(5-1)^2+(1+3)^2=4^2+4^2=32>25,

so the point is outside the circle.

When the answer choices are points, plotting them gives you a quick first look. But a point can sit a tiny bit off the circle and still look like it's on it. So if a point looks close, plug it in before you decide. The graph shows you where to look, but only the substitution shows that the two sides are exactly equal.

Check your understanding:

Is $(2,5)$ inside, on, or outside the circle $(x+2)^2+(y-1)^2=36$ ?

Common mistake:

Calling any point inside the circle a point “on” the circle. With its equals sign, the equation draws only the edge. A point is on the circle only when plugging it in makes the two sides equal, the way $(4,1)$ gives exactly $25$ .

Calculator loads as you approach

Plotting gives you a quick look. Plugging in settles it.

Add the coordinate condition as a second graph

Turn the clue into a graph

Many SAT questions tell you something about a point on a circle without giving you both of its coordinates. Turn that clue into a second graph. The point you want is where the two graphs meet.

What the question says	What to graph
the point has $x=c$	`x=c`
the point has $y=c$	`y=c`
the point is $(a,a+d)$	`y=x+d`
the point is on the line $y=mx+b$	`y=mx+b`

For example, graph

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

and the condition

x=5.

The graphs meet at

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

The number of meeting points tells you something too:

Two: the line cuts through the circle.
One: the line is tangent to the circle.
Zero: the line misses the circle completely.

Before you decide a line misses, make sure both graphs are actually showing. And if the line comes close to the circle, zoom in. A line that barely touches or barely misses is easy to misjudge.

Try it yourself:

Change x=5 to x=7. How many times do you think the line meets the circle now? Make your guess, then check it on the graph.

Check your understanding:

For the circle $(x-2)^2+(y+1)^2=25$ , what is the greater $y$ -coordinate of a point whose $x$ -coordinate is $5$ ?

Common mistake:

Grabbing the wrong number from the point you clicked. If the point is $(5,3)$ and the question asks for the $y$ -coordinate, the answer is $3$ , not $5$ .

Calculator loads as you approach

One x-value can give two points on a circle. Check which y-value the question wants.

Example: Turn a point pattern into a line

Worked example

The graph of

x^2+y^2-8x+6y+9=0

is a circle in the $xy$ -plane. The point $(a,a-5)$ lies on the circle, and $a>3$ . What is the value of $a$ ?

A
$3+\sqrt7$
B
$3+\dfrac{\sqrt7}{2}$
C
$4+\sqrt7$
D
$5+\sqrt7$

Step 1

Turn the point pattern into a line

Look at $(a,a-5)$ . Whatever $a$ is, the second coordinate is $5$ less than the first. Points like $(7,2)$ and $(0,-5)$ fit that pattern, and every point that fits it lies on the line

y=x-5.

So the point you want is where this line meets the circle, and its $x$ -coordinate is $a$ .

Step 2

Graph both conditions

Graph the circle equation and y=x-5. They meet twice, at $x$ -coordinates of about

0.354 \qquad\text{and}\qquad 5.646.

The question says $a>3$ , so $a$ is the greater one, about $5.646$ .

Calculator loads as you approach

The x-coordinates where the graphs meet are the possible values of a.

Step 3

Match the exact value

Desmos gave a decimal, but the choices are exact. Since $\sqrt7\approx2.646$ , choice A, $3+\sqrt7$ , is the one near $5.646$ . To be sure, let's find the exact value.

Put $x=a$ and $y=a-5$ into the circle equation:

a^2+(a-5)^2-8a+6(a-5)+9=0.

Expand $(a-5)^2=a^2-10a+25$ and $6(a-5)=6a-30$ , then combine like terms:

2a^2-12a+4=0.

Divide both sides by $2$ :

a^2-6a+2=0.

The quadratic formula gives

a=\frac{6\pm\sqrt{36-8}}{2}=\frac{6\pm2\sqrt7}{2}=3\pm\sqrt7.

Since $a>3$ ,

a=3+\sqrt7.

The answer is A.

Finish the solution

Use the center and radius to find the circle's top and bottom, then let the condition $k>0$ choose between them.

Find the positive horizontal tangent

Finish the solution

The horizontal line $y=k$ is tangent to the circle

(x-5)^2+(y+4)^2=25.

If $k>0$ , what is the value of $k$ ?

First steps

Find the circle's center and radius.
Find its greatest and least $y$ -coordinates.
Each of those gives a horizontal tangent line $y=k$ .
Keep the one with $k>0$ .

Finish it

Your answer

Calculator loads as you approach

Add a horizontal line, or select the circle to see its top and bottom points.

Practice problems

Now it's your turn. Give each problem a real try before you open its hint.

Write a circle equation

Practice problem

A circular signal boundary in the $xy$ -plane has center $(-5,2)$ and radius $8$ . Which equation represents the boundary?

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Watch the signs, and square the radius.

Read the diameter from an expanded equation

Practice problem

Coordinates are measured in meters. A circular boundary is represented by

x^2+y^2+14x-10y+10=0.

What is the diameter of the circle?

Calculator loads as you approach

Graph the whole equation, then use its highest and lowest points.

Test which point lies on a circle

Practice problem

Which point lies on the circle

(x-1)^2+(y+3)^2=25?

Calculator loads as you approach

Plot the choices for a quick look, then plug in to be sure.

Use a tangency point

Practice problem

The line

y=2x+14

is tangent to the circle

(x+1)^2+(y-2)^2=20

at the point $(a,b)$ . What is the value of $a+b$ ?

Your answer

Calculator loads as you approach

Graph both and click the one point where they touch. You’ll need both of its coordinates.

Finish the lesson

5 practice examples left

Finish the remaining questions correctly to complete this lesson.

Quick recap

Type a whole circle equation into Desmos as written. Don't solve it for $y$ first.
In $(x-h)^2+(y-k)^2=r^2$ , the center is $(h,k)$ and the radius is the positive number $r$ . Watch the flipped signs.
A circle reaches $r$ units left, right, up, and down from its center.
Plug a point in to see whether it's inside, on, or outside the circle.
Turn an extra condition into a second graph, then click where the graphs meet.
Two meeting points mean the line crosses the circle, one means it's tangent, and zero means they don't meet.
Desmos may give a decimal. Use the question, the answer choices, or exact algebra to give the answer in the form asked for.
Read clean center-radius form by hand. Graph when the equation is expanded or an extra condition is easier to see.

Next lesson

Custom regression for one unknown parameter

Keep a known model exactly as given while Desmos finds one missing constant from one fact about it.

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Practice

Practice this lesson

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

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