Build circles from coordinate information

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Difficulty: Intermediate
Estimated time: 33 minutes
Domains: Geometry and Trigonometry
Techniques: Diameter-endpointsInscribed-right-angleMidpointDistanceConstrained-centerTwo-circle-tangency

What you’ll learn

Tell when the coordinates hide the center or radius.
Spot a diameter across from a right angle on the circle.
Find the center and $r^2$ from the ends of a diameter.
Use Desmos midpoint and distance when a diameter’s coordinates are messy.
Find the missing end of a diameter from the center.
Find a center from a rule, like a line it sits on, and two points on the circle.
Match the center distance of two tangent circles to the sum or difference of their radii.
Write the finished circle in standard form.

Prerequisites

Why this matters on the SAT

Turn two endpoints into a complete circle

Instead of a center and a radius, the SAT may give you the two ends of a diameter, a segment that crosses the circle through its center. Once you know you have a diameter, the rest follows. Its midpoint is the center, and half its length is the radius. Sometimes the word itself never appears, and a right angle drawn on the circle gives the diameter away instead.

Let the numbers pick how you work. When the midpoint and the changes come out as clean whole numbers, do it by hand. When the coordinates are messy fractions or decimals, the arithmetic is easy to slip on, so enter the two endpoints in Desmos and use midpoint(A,B) and (distance(A,B)/2)^2.

Solution to the example

Start with the center. It’s the midpoint of $A$ and $B$ , so average the $x$ -coordinates and average the $y$ -coordinates:

\left(\frac{-3+5}{2},\frac{-2+4}{2}\right)=(1,1).

Now the radius. From $A$ to $B$ you go $8$ units right and $6$ units up, so the squared diameter is

8^2+6^2=100.

The radius is half the diameter, and halving a length divides its square by $4$ . So $r^2$ is one-fourth of the squared diameter:

r^2=\frac{100}{4}=25.

The equation is

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

so the answer is B.

SAT example

Segment

AB

is the given diameter.

In the $xy$ -plane, the endpoints of a diameter of a circle are $A(-3,-2)$ and $B(5,4)$ . Which equation represents the circle?

A
$(x+1)^2+(y+1)^2=25$
B
$(x-1)^2+(y-1)^2=25$
C
$(x-1)^2+(y-1)^2=100$
D
$(x-3)^2+(y-2)^2=25$

Recognize a hidden diameter

An inscribed angle is an angle whose vertex sits on the circle. In the diagram, all three points $A$ , $B$ , and $C$ lie on the circle, and $\angle ACB$ is a right angle.

The chord across from the right angle at

C

runs straight through the center

O

That’s no accident. Whenever three points sit on a circle and the angle at one of them is a right angle, the chord joining the other two is a diameter. We’ll call this the inscribed-right-angle signal. Here the right angle is at $C$ , so $\overline{AB}$ is a diameter, even though nobody called it one. From here, $A$ and $B$ are ordinary diameter ends.

Check your understanding:

Points $P$ , $Q$ , and $R$ all lie on a circle, and $\angle PRQ=90^\circ$ . The coordinates of $P$ and $Q$ are messy decimals. Which segment is the diameter, and what would you type into Desmos next?

Build the circle in four decisions

These questions get much easier when you build the circle first and write the equation last. Answer four questions in order. The SAT example shows each one.

What do the coordinates give you? There, $A$ and $B$ were the ends of a diameter. Other questions give a center and one point on the circle, an axis or another circle it’s tangent to, or a line the center sits on.
Where is the center? For a diameter, it’s the midpoint, $(1,1)$ in the example. If the question only tells you facts about the center, like a line it sits on, combine them with its other conditions, such as points on the circle.
What is $r^2$ ? Square the horizontal and vertical changes from the center to any point on the circle, and add them. From $(1,1)$ to $B(5,4)$ , that’s $4^2+3^2=25$ .
What is the equation? Put the center and $r^2$ into standard form: $(x-1)^2+(y-1)^2=25$ .

Check your understanding:

A problem gives you the center and one point on the circle. Which of the four questions are already answered, and what do you work out next?

Make diameter endpoints do two jobs

Say $A(x_1,y_1)$ and $B(x_2,y_2)$ are the ends of a diameter. Those two points give you everything you need.

Job 1: find the center

The center sits exactly halfway along the diameter, so it’s the midpoint:

(h,k)= \left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right).

Job 2: find the squared radius

First get the squared length of the diameter:

AB^2=(x_2-x_1)^2+(y_2-y_1)^2.

The radius is half the diameter, $r=\frac{AB}{2}$ , so

r^2=\frac{AB^2}{4} = \frac{(x_2-x_1)^2+(y_2-y_1)^2}{4}.

Working with $r^2$ the whole way means you never take a square root only to square it again.

Common mistake:

The distance between the endpoints is the diameter, not the radius. Use it as the radius and your circle comes out twice as big. Divide the distance by $2$ , or divide the squared distance by $4$ when the equation needs $r^2$ . That’s the trap in choice C of the first example: $100$ is the squared diameter, not $r^2$ .

Check your understanding:

The ends of a diameter are $10$ units apart horizontally and $4$ units apart vertically. What is $r^2$ ?

Use Desmos midpoint and distance for messy coordinates

A graph can’t tell you that $A$ and $B$ are the ends of a diameter. The question has to say so, or a circle fact like the inscribed-right-angle signal has to show it. Once you know, Desmos can do the arithmetic:

A=(-7/2,5/4)
B=(13/2,29/4)
midpoint(A,B)
distance(A,B)/2
(distance(A,B)/2)^2

midpoint(A,B) gives the center, $\left(\frac32,\frac{17}{4}\right)$ . The radius line shows about $5.831$ . Don’t round it and square it: $5.831^2$ is about $34.0006$ , which isn’t exact. Whenever Desmos shows a decimal radius, square the half-distance there, as the last line does, or use the squared-difference formula above. Here that gives exactly $r^2=34$ . Put the center and $r^2$ into standard form:

\left(x-\frac32\right)^2+\left(y-\frac{17}{4}\right)^2=34.

For more circles built this way, see Circle geometry with distance and midpoint.

Calculator loads as you approach

The midpoint is the center. Squaring the half-distance gives the exact r² = 34, with no rounding.

Recover the other endpoint

Sometimes you know the center and one end of a diameter, and the question wants the other end. The center is still the midpoint, exactly halfway between the two ends. So the step that takes you from one end to the center also takes you from the center to the other end.

Say one end is $A=(-4,7)$ and the center is $C=(1,3)$ . The step from $A$ to $C$ is

(1-(-4),3-7)=(5,-4),

which is $5$ right and $4$ down. Take the same step again from the center:

B=(1+5,3-4)=(6,-1).

To check, average the two ends: $\left(\frac{-4+6}{2},\frac{7+(-1)}{2}\right)=(1,3)$ , the center.

Check your understanding:

The center is $(2,5)$ , and one end of a diameter is $(-1,9)$ . What’s the other end?

Tangent circles come down to center distance

A circle is tangent to a line or to another circle when it touches it at exactly one point. The simplest case is an axis. A circle centered at $(3,5)$ that’s tangent to the $y$ -axis touches it at $(0,5)$ , so its radius is the center’s distance to that axis, $3$ . If it were tangent to the $x$ -axis instead, its radius would be $5$ .

For two tangent circles, the distance that matters is the one between their centers. The centers and the touchpoint sit on one line, so that distance is the two radii, added or subtracted.

Outside each other, the radii add along the line through the centers. One inside the other, they subtract, because the bigger radius contains the smaller one.

Here’s each case with numbers. Take centers $(1,2)$ and $(9,2)$ with radii $3$ and $5$ . The centers are $8$ apart, which is $3+5$ , so the circles touch from the outside. Now take centers $(1,2)$ and $(5,2)$ with radii $7$ and $3$ . The centers are $4$ apart, which is $7-3$ , so the smaller circle sits inside the bigger one and touches it.

In symbols, call the distance between the centers $d$ and the radii $r_1$ and $r_2$ .

If the circles sit outside each other, they’re externally tangent, and $d=r_1+r_2$ .
If one circle sits inside the other, they’re internally tangent, and $d=\lvert r_1-r_2\rvert$ , the bigger radius minus the smaller.

To get $d$ from coordinates, square the horizontal and vertical changes between the centers and add them. That gives $d^2$ , so you can skip the square root: compare $d^2$ with $(r_1+r_2)^2$ or $(r_1-r_2)^2$ , using exact numbers.

When a question says only that two circles are tangent, other details pick the relationship. Words like lies entirely inside mean internal tangency. Separate circles that touch from the outside mean external tangency.

Check your understanding:

Two circles have centers $(-2,1)$ and $(4,9)$ and radii $6$ and $4$ . Are they externally tangent, internally tangent, or neither?

Common mistake:

Adding the radii works only when the circles touch from the outside. So before you add, ask whether one circle sits inside the other. If it does, subtract. Then check your answer against every other condition the question gives, using exact numbers, not how the graph looks.

Example: Build a center constrained to a line

Worked example

The two given points lie on the circle, and its center lies somewhere on the line.

A circle in the $xy$ -plane passes through points $P(0,2)$ and $Q(4,4)$ . Its center lies on the line $y=x-2$ . Which equation represents the circle?

A
$(x-2)^2+(y-3)^2=5$
B
$(x-3)^2+(y+1)^2=10$
C
$(x-3)^2+(y-1)^2=10$
D
$(x-3)^2+(y-1)^2=20$

Step 1

Write every possible center

This is the trickiest setup yet, because no diameter hands you the center. The line is your way in. Call the center’s $x$ -coordinate $t$ . The center is on $y=x-2$ , so its $y$ -coordinate is $t-2$ :

C=(t,t-2).

Now the center has one unknown instead of two.

Step 2

Use equal radii

$P$ and $Q$ are both on the circle, so they’re the same distance from the center: $CP=CQ$ . From $P(0,2)$ , the changes to the center are $t$ and $t-4$ . From $Q(4,4)$ , they’re $t-4$ and $t-6$ . Set the squared distances equal:

t^2+(t-4)^2=(t-4)^2+(t-6)^2.

The $(t-4)^2$ on each side cancels:

t^2=(t-6)^2.

Step 3

Find the center

Expand only what’s left:

\begin{aligned} t^2&=t^2-12t+36\\[1.4em] 12t&=36\\[1.4em] t&=3. \end{aligned}

So the center is

(t,t-2)=(3,1).

Step 4

Find the radius and write the equation

Use the center and point $P$ :

r^2=(0-3)^2+(2-1)^2=9+1=10.

The circle is

(x-3)^2+(y-1)^2=10.

Check it with point $Q$ :

(4-3)^2+(4-1)^2=1+9=10.

It fits, so the answer is C.

Common mistake:

Why not use the midpoint of $P$ and $Q$ as the center? That only works if $\overline{PQ}$ is a diameter, and the question says only that both points are on the circle. That midpoint, $(2,3)$ , gives choice A, and it isn’t even on the line $y=x-2$ . Use equal distances from the center to each point, together with the line the center sits on.

Practice problems

Problem 1 is one for Desmos, and Problems 2 and 3 work out by hand. The last one puts tangency to an axis together with one circle inside another.

Use midpoint and distance

Practice problem

In the $xy$ -plane, the endpoints of a diameter of a circle are

A(-7.38,2.46) \quad\text{and}\quad B(12.94,15.72).

Which equation represents the circle?

Calculator loads as you approach

Define A and B as points first. Then if you fix a typo in a point, the midpoint and distance results update with it.

Recover the opposite endpoint

Practice problem

A circle has center $(2,-1)$ . One endpoint of a diameter is $(-3,5)$ , and the other endpoint is $(a,b)$ . What is the value of $a+b$ ?

Your answer

Calculator loads as you approach

The center is the midpoint. Plot all three points if you want to check that they line up, evenly spaced.

Combine axis and circle tangency

Practice problem

A circle $C$ in the $xy$ -plane has equation

(x-2)^2+y^2=36.

Circle $D$ has center $(h,0)$ , where $h>0$ . Circle $D$ is tangent to the $y$ -axis, lies entirely inside circle $C$ , and is tangent to circle $C$ . Which equation represents circle $D$ ?

Calculator loads as you approach

Solve with exact center distances. Graph your finished circles to check them, but looking tangent on a graph doesn’t prove they’re tangent.

Finish the lesson

3 practice examples left

Finish the remaining questions correctly to complete this lesson.

Quick recap

Ask what the coordinates stand for before you use a formula.
A right angle with its vertex on the circle sits across from a diameter.
The midpoint of a diameter is the center, and half its length is the radius.
Get $r^2$ straight from the ends of a diameter: one-fourth of their squared distance.
To find a missing end, take the step from the known end to the center once more.
A center on a line fits that line and is equally far from every given point on the circle.
When a circle is tangent to an axis, its radius is the center’s distance to that axis.
For tangent circles, the center distance is the sum of the radii from outside and the absolute difference with one inside.
Write standard form last, once the center and $r^2$ are settled.
Work clean whole numbers by hand. For messy diameter ends, use midpoint(A,B) for the center and (distance(A,B)/2)^2 for an exact $r^2$ .

Next lesson

Circle geometry with distance and midpoint

Use Desmos midpoint and distance to turn circle coordinates into centers, radii, and equations.

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Practice

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

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