Circle equations in the graph

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Difficulty: Intermediate
Estimated time: 55 minutes
Domains: Geometry and Trigonometry
Techniques: CirclesGraphingCircle-equationsCenter-radius-formPoint-membershipIntersectionsTangency

What you’ll learn

Graph a complete circle equation without solving it for y.
Read the center and radius from center-radius form.
Use the center and radius to find the circle’s coordinate extent.
Classify a point as inside, on, or outside a circle.
Translate a coordinate condition into a second graph.
Read circle-line intersections and recognize tangency.
Choose when graphing is useful and when hand reasoning is faster.

Prerequisites

Why this matters on the SAT

Turn a circle equation into a picture

SAT circle questions can hide useful geometry inside an equation. Desmos can graph the complete equation as a circle, reveal its highest and lowest points, and show where it meets a line or another condition.

That picture is especially valuable when the equation is expanded and the center is not obvious.

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?

A
$-12$
B
$-4$
C
$4$
D
$12$

Fast Desmos solution

Enter the circle equation on one line and y=9 on another. Click the two intersections. Desmos shows

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

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

Desmos found the points, but the prompt decided which coordinate to report. The calculator does not know that the negative intersection must be rejected.

Calculator loads as you approach

Click the two intersections, then return to the prompt and select the positive x-coordinate.

When should you use Desmos?

Choose the method that makes the circle’s useful information easiest to see.

Graph the circle when…

the equation is expanded and its location is not immediately clear;
the question asks for a possible $x$ - or $y$ -coordinate;
you need to test whether a point lies on the circle;
a line or coordinate condition creates useful intersections; or
the answer depends on zero, one, or two intersections.

Use hand reasoning first when…

the equation is already in clean center-radius form;
the question asks only for the center, radius, diameter, or coordinate range;
one short substitution tests a point exactly; or
exact algebra is needed after Desmos shows only a decimal.

Read clean center-radius form before graphing. Use the graph when it makes an expanded equation or an added condition easier to see.

Use the later circle geometry with distance and midpoint lesson when the problem gives diameter endpoints. Recovering a circle from three geometric points belongs in geometry to coordinates to circle regression.

Check your understanding:

Which is the stronger reason to begin with Desmos: finding the radius of $(x-2)^2+(y+5)^2=36$ , or finding where $x^2+y^2-6x+4y-12=0$ intersects $y=2x-1$ ?

1. Enter the complete circle equation

Graph a relation, not a function

A circle is usually not one function of $x$ . Most $x$ -coordinates inside the circle’s horizontal range pair with two $y$ -coordinates, one above the center and one below it.

Desmos can still graph the complete relation. Enter the equation exactly as written:

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

Do not place y= before the complete equation.

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

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

These points form the vertical diameter. The center is halfway between their $y$ -coordinates, at $(2,-1)$ , and the radius is $5$ .

Desmos does not reliably mark the leftmost and rightmost points as clickable extrema. When the equation is in center-radius form, derive those coordinates from the center and radius instead of estimating them from the grid.

Try it yourself:

Replace the equation with x^2+y^2+2x-8y-8=0. Select the circle and predict its highest and lowest $y$ -coordinates before clicking them.

Check your understanding:

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

Common mistake:

Typing y=x^2+y^2-4x+2y-20=0. The original equation already contains both graphing variables. Adding another y= changes the relation instead of graphing the given circle.

Calculator loads as you approach

Select the circle to reveal its highest and lowest points.

2. Read center-radius form before graphing

The signs identify the center

The center-radius form of a circle is

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

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

For example,

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

can be written as

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

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

Use the radius to find every extreme

Quantity	Value
least $x$	$h-r$
greatest $x$	$h+r$
least $y$	$k-r$
greatest $y$	$k+r$

For the circle above:

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

This hand calculation is faster and more exact than tracing the leftmost and rightmost points.

Try it yourself:

Change the right side from $49$ to $81$ . Predict the new radius and all four extreme coordinates, then check the new circle.

Check your understanding:

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

Common mistake:

Reporting $49$ as the radius of $(x+4)^2+(y-3)^2=49$ . The right side is $r^2$ , so the positive 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.

3. Test a point against the circle

Compare the point with the boundary value

For

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

substitute a point $(a,b)$ into the left side:

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

Then compare the result with $r^2$ .

Comparison	Location
less than $r^2$	inside the circle
equal to $r^2$	on the circle
greater than $r^2$	outside the circle

Consider

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

For $(4,1)$ :

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

so the point lies on the circle.

For $(5,1)$ :

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

so the point lies outside the circle.

Graphing a candidate point is useful when answer choices are coordinates. If the point appears close to the circle, substitute it into the equation before deciding. The graph is a check, not a proof of an exact equality.

Check your understanding:

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

Common mistake:

Treating every point in the interior as a point on the circle. A circle equation with an equals sign graphs only the boundary. A point is on the circle only when substitution makes the two sides equal.

Calculator loads as you approach

Plotting gives a fast visual check. Substitution confirms the exact classification.

4. Add the coordinate condition as a second graph

Translate the condition, then click the intersections

Many SAT questions describe a point on a circle without giving both coordinates directly. Turn the extra condition into a second graph:

Condition	Second 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 lies on $y=mx+b$	`y=mx+b`

For example, graph

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

and the condition

x=5.

The graphs intersect at

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

The number of intersections also has geometric meaning:

two intersections: the line crosses the circle;
one intersection: the line is tangent to the circle; and
zero intersections: the line does not meet the circle.

Before concluding that there are no intersections, confirm that both graphs are visible and inspect a near-tangent case closely.

Try it yourself:

Change x=5 to x=7. Predict the number of intersections first, then click the intersection to test your prediction.

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:

Reporting the wrong coordinate from an intersection. If the point is $(5,3)$ and the question asks for the $y$ -coordinate, submit $3$ , not $5$ .

Calculator loads as you approach

The same x-coordinate can produce two points on a circle. Read the requested y-coordinate carefully.

5. Example: Translate a parameterized point

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.

If $a>3$ , 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

In $(a,a-5)$ , the second coordinate is $5$ less than the first. Every point with that pattern lies on

y=x-5.

At an intersection with the circle, the $x$ -coordinate represents $a$ .

Step 2

Graph both conditions

Graph the circle equation and y=x-5. Desmos shows two intersections with approximate $x$ -coordinates

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

The condition $a>3$ keeps the greater value.

Calculator loads as you approach

The intersection x-coordinates are the possible values of a.

Step 3

Match the exact value

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

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

Simplifying gives

a^2-6a+2=0.

The quadratic formula gives

a=3\pm\sqrt7.

Since $a>3$ ,

a=3+\sqrt7.

The answer is A.

Finish the solution

Use the center and radius to identify the horizontal tangent that satisfies the remaining condition.

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

Identify the circle’s center and radius.
Find its greatest and least $y$ -coordinates.
Each vertical extreme determines a horizontal tangent line $y=k$ .
Keep the value that satisfies $k>0$ .

Finish it

Your answer

Calculator loads as you approach

Add a horizontal line or select the circle and inspect its top and bottom points.

Practice problems

SAT practice problems

Read clean center-radius form before graphing. Use Desmos when the equation is expanded or when a point or line condition creates useful intersections.

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?

Calculator loads as you approach

Check the signs in the center 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 complete equation and use the highest and lowest points to read the vertical diameter.

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 candidate points, then substitute to verify the exact match.

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

A tangent line and circle have one intersection. Read both coordinates before evaluating the requested sum.

Finish the lesson

5 practice examples left

Finish the remaining questions correctly to complete this lesson.

Quick recap

Enter a complete circle equation directly. Do not 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 value $r$ .
The circle extends $r$ units left, right, up, and down from its center.
Substitute a point to determine whether it lies inside, on, or outside the circle.
Turn an extra coordinate condition into a second graph, then click the circle-line intersections.
Two intersections mean the line crosses the circle, one means tangency, and zero means the graphs do not meet.
Desmos may show decimal intersections. Use the prompt, answer choices, or exact algebra to report the required form.
Read clean center-radius form by hand. Use the graph when it makes an expanded equation or an added condition easier to see.

Next lesson

Custom regression for one unknown parameter

Keep a known model intact while Desmos recovers one missing constant from a numerical condition.

Start next lesson

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