Circle geometry with distance and midpoint

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

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

What you’ll learn

Name coordinate points in Desmos and use distance and midpoint on them.
Turn the two ends of a diameter into a circle's center, radius, $r^2$ and equation.
Use the center and one point on the circle to find the radius or the area.
Spot when a right angle inscribed in a circle makes the side across from it a diameter.
Keep values exact instead of rounding early, and check them.
Tell when these commands help and when working by hand is faster.

Prerequisites

Circle equations in the graph0%
Read a circle’s center and radius and write center-radius form.

Why this matters on the SAT

Turn two points into a complete circle

SAT circle questions often give you points instead of an equation. When two points are the ends of a diameter, two short Desmos commands turn them into the circle's center and radius. From there you can write the equation, find the area, or find another point on that diameter.

SAT example

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

Fast Desmos solution

Name the endpoints $A$ and $B$ , then type:

midpoint(A,B)
distance(A,B)/2

Desmos gives the center, $(-1,1)$ , and the radius, $5$ . Put them into center-radius form,

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

and the circle is

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

The answer is A.

Desmos does the arithmetic. Your part is noticing that the two points are the ends of a diameter, and then giving exactly what the question asks for. So before you type a command, ask yourself one thing: diameter or radius?

Calculator loads as you approach

The two endpoints, the center and radius Desmos finds, and the circle they make.

When should you use this method?

These commands pay off when a circle question gives you points and the arithmetic isn’t quick. When it is quick, your head is faster.

Reach for distance or midpoint

The question gives the two ends of a diameter, like $(2,-3)$ and $(-4,5)$ in the example above.
You know the center and one point on the circle, like a stage centered at $(2,-1)$ that passes through $(8,5)$ .
A triangle has all three corners on the circle and a right angle, so one of its sides is a hidden diameter.
The coordinates are awkward, like $(-7,13)$ and $(4,-9)$ , and the subtraction would be easy to get wrong in your head.

Work it out by hand first

The points share an $x$ - or $y$ -coordinate, so the distance is one subtraction. For example, $(-2,4)$ and $(5,4)$ are $7$ apart.
You can see the midpoint right away, like $(2,3)$ halfway between $(1,1)$ and $(3,5)$ .
The equation is already in center-radius form, like $(x-4)^2+(y+2)^2=9$ , which shows the center $(4,-2)$ and the radius $3$ .
The question is really about arcs, sectors, tangent lines or another circle fact, not the distance between coordinates.

Two points on a circle aren’t automatically the ends of a diameter. The question, or a circle fact, has to tell you they are.

Have three points on the circle and no diameter? Save those for geometry to coordinates to circle regression.

Check your understanding:

A circle has center $(3,-1)$ and passes through $(3,6)$ . Would you start with the distance command or with mental math? What’s the radius?

Name the points, then use the commands

Name each point once

Say you have

A=(-2,1) \qquad\text{and}\qquad B=(4,9).

Type each point on its own line:

A=(-2,1)
B=(4,9)

Then type the two commands:

midpoint(A,B)
distance(A,B)

Desmos returns

\operatorname{midpoint}(A,B)=(1,5)

and

\operatorname{distance}(A,B)=10.

Notice the two kinds of answer. The midpoint is a point, so it has two coordinates. The distance is a length, so it's one number, and it's never negative.

The order of the points doesn't matter. Swap $A$ and $B$ and you get the same midpoint and the same distance:

\operatorname{midpoint}(A,B)=\operatorname{midpoint}(B,A)

and

\operatorname{distance}(A,B)=\operatorname{distance}(B,A).

Try it yourself:

Change $B$ from $(4,9)$ to $(4,-7)$ . Before you look, predict the new midpoint. Will the distance change? Check your guesses, then change $B$ back.

Check your understanding:

Say midpoint(P,Q) returns $(-3,4)$ and distance(P,Q) returns $12$ . Which output is a point, and which is a length? If $\overline{PQ}$ were a diameter, what would the radius be?

Common mistake:

Type each point exactly as the question gives it, with $x$ first and $y$ second. If you retype the coordinates inside every command, it’s easy to flip a pair or drop a minus sign in one of them. Naming the points once means both commands use the same pairs.

Calculator loads as you approach

Both commands use the named points A and B.

Turn diameter endpoints into a circle

Midpoint gives the center

A diameter passes through the center, and the center sits exactly halfway along it. So if $A=(x_1,y_1)$ and $B=(x_2,y_2)$ are the ends of a diameter, the center $C$ is their midpoint:

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

In Desmos, type:

C=midpoint(A,B)

Distance gives the diameter, not the radius

The distance from one end of a diameter to the other runs all the way across the circle, so it's the diameter:

d=\operatorname{distance}(A,B).

The radius is half of that:

r=\frac d2.

In Desmos, type:

R=distance(A,B)/2

Use a capital $R$ . Desmos keeps a lowercase $r$ for polar graphs, so r=distance(A,B)/2 would also draw a circle around the origin.

If the equation or the area needs $r^2$ , let Desmos do the squaring:

R^2

(distance(A,B)/2)^2

That way you don't have to round an irrational radius, like $3\sqrt2$ , whose decimal never ends, before you square it.

Let's try it with diameter endpoints

A=(-3,2) \qquad\text{and}\qquad B=(5,8).

The midpoint is

C=(1,5).

The distance between the ends is

\sqrt{(5-(-3))^2+(8-2)^2} =\sqrt{8^2+6^2} =10,

so the radius is $5$ . Put the center and the radius into center-radius form, and the circle is

(x-1)^2+(y-5)^2=25.

Check your understanding:

A diameter has endpoints $(1,2)$ and $(7,8)$ . Find the center and $r^2$ without rounding.

Common mistake:

In the example above, the ends are $10$ apart. Use $10$ as the radius and you’d write $=100$ instead of $=25$ , a circle twice as wide as the real one. Halve the distance first, or divide the squared distance by $4$ when you need $r^2$ .

Calculator loads as you approach

C = (1, 5) is the center, and R = 5 is half the distance from A to B.

Use the center and one point on the circle

Their distance is one radius

If $C$ is the center and $P$ is any point on the circle, then

r=\operatorname{distance}(C,P).

No halving this time. $\overline{CP}$ goes from the center out to the circle, which is one radius, not the whole way across.

Worked example

A circular stage has center

C=(2,-1)

and passes through

P=(8,5).

What is its area in terms of $\pi$ ?

Square the coordinate changes and add them to get $r^2$ :

\begin{aligned} r^2 &=(8-2)^2+(5-(-1))^2\\[1.4em] &=6^2+6^2\\[1.4em] &=72. \end{aligned}

So the area is

\text{area}=\pi r^2=\boxed{72\pi}.

Here's the tricky part in Desmos. The radius shows up as a decimal, $8.485\ldots$ , because $r=6\sqrt2$ . The area needs $r^2$ , not $r$ . Round the radius to $8.49$ and square it, and you get $72.0801$ : close, but not exact. Name the radius R=distance(C,P) and type R^2 instead, and Desmos shows $72$ without you rounding anything. Desmos still works in decimals, though, so check the exact value from the coordinate changes:

r^2=6^2+6^2=72.

Try it yourself:

Change $P$ to $(5,3)$ . Predict $r^2$ and the number in front of $\pi$ in the area, then check both in the calculator.

Check your understanding:

A circle has center $(-4,3)$ and passes through $(1,15)$ . What are the radius and $r^2$ ?

Common mistake:

Divide by $2$ only when the two points are opposite ends of a diameter. Halve the stage’s radius by mistake and you’d get an area of $18\pi$ , only a quarter of the real $72\pi$ .

Calculator loads as you approach

The radius shows as a long decimal, but

R^2

shows 72, the number the equation and the area need.

Let a right angle reveal the diameter

The side across from the right angle spans the circle

Say a triangle is inscribed in a circle, meaning all three of its corners sit on the circle. If one of its angles is $90^\circ$ , the side across from that angle is a diameter.

For triangle $ABC$ with all three corners on the circle, if $\angle ACB=90^\circ$ , then $\overline{AB}$ is a diameter.

So a right-angle question turns into a diameter question, and you already know how to handle those. You don't even need the third corner to find the center.

Check your understanding:

Triangle $PQR$ is inscribed in a circle, and $\angle Q=90^\circ$ . Which side is a diameter?

Example: Find a circle from an inscribed right angle

Worked example

In the $xy$ -plane, triangle $ABC$ is inscribed in a circle. The vertices are

A=(-2,1),\qquad B=(6,7),\qquad C=(5,0),

and $\angle ACB$ is a right angle. Which equation represents the circle?

A
$(x-2)^2+(y-4)^2=25$
B
$(x+2)^2+(y+4)^2=25$
C
$(x-2)^2+(y-4)^2=100$
D
$(x-5)^2+y^2=25$

Step 1

Find the diameter

The right angle is at $C$ , so the side across from it, $\overline{AB}$ , is a diameter.

You won't need $C$ to find the circle. It tells you where the triangle sits, but the diameter's ends, $A$ and $B$ , are enough.

Step 2

Find the center and radius

Type $A$ and $B$ , then:

midpoint(A,B)
distance(A,B)/2

Desmos returns

\operatorname{midpoint}(A,B)=(2,4)

and

\frac{\operatorname{distance}(A,B)}{2}=5.

Calculator loads as you approach

AB, the side across from the right angle, is the diameter of the circle.

Step 3

Write the equation

With center $(2,4)$ and radius $5$ , center-radius form gives

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

That's

(x-2)^2+(y-4)^2=25,

so the answer is A.

As a check, $C=(5,0)$ should be on this circle too, and it is:

(5-2)^2+(0-4)^2=3^2+(-4)^2=25.

Common mistake:

In a right triangle, the side across from the $90^\circ$ angle is the hypotenuse, and that’s the diameter. The other two sides, the legs, meet at the right angle and are shorter than the hypotenuse, so neither one can be a diameter. Find the hypotenuse before you type any coordinates.

Keep it exact and answer the question

A Desmos output is a step along the way, not your final answer. Before you submit:

Check whether your distance is a radius or a diameter.
Reread what the question wants: $r$ , $r^2$ , an equation, an area, a coordinate, or an expression built from coordinates.
Don't round until the very last step.
Check the geometry with a midpoint, equal distances from the center, or a point plugged into your equation.

Two exact formulas for $r^2$ help here. For the ends of a diameter:

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

For a center $(h,k)$ and a point $P$ on the circle:

r^2=(x_P-h)^2+(y_P-k)^2

When a multiple-choice question has answers with square roots or fractions, use the Desmos decimal to spot the likely choice. Then check that choice exactly before you pick it.

Check your understanding:

The distance between two diameter endpoints is $6\sqrt5$ . The question asks for the area of the circle in the form $k\pi$ . What value of $k$ do you submit?

Finish the solution

The endpoints and the center are already in the calculator. Finish the radius and the area, then submit the number in front of $\pi$ .

Finish the solution

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

A=(-4,6) \qquad\text{and}\qquad B=(8,-2).

The area of the circle is $k\pi$ square units. What is the value of $k$ ?

First steps

The center is midpoint(A,B), and Desmos shows $(2,2)$ .
Find the diameter with distance(A,B).
Halve the diameter to get the radius.
Square the radius. That's $k=r^2$ .

Finish it

Your answer

Calculator loads as you approach

Add lines for the diameter, the radius and

r^2

Practice problems

Your turn. Before you type anything, ask what the two points are: the ends of a diameter, or a center and a point on the circle?

Practice problem

In the $xy$ -plane, a circle has a diameter with endpoints $(2,-3)$ and $(8,9)$ . Which equation represents the circle?

Calculator loads as you approach

The ends of a diameter give you the center and the radius.

Practice problem

A circular stage in the $xy$ -plane has center $(1,-2)$ and passes through $(5,4)$ . Its area is $k\pi$ square units. What is the value of $k$ ?

Your answer

Calculator loads as you approach

A center and a point on the circle give you the radius directly.

Practice problem

A circular garden in the $xy$ -plane has center $(-5,2)$ and passes through $(-5,11)$ . Its area is $k\pi$ square units. What is the value of $k$ ?

Your answer

Calculator loads as you approach

The shared

x

-coordinate makes this one quick by hand. The calculator is here if you want to check

9^2

Practice problem

Triangle $ABC$ is inscribed in a circle in the $xy$ -plane. The vertices are

A=(-5,1),\qquad B=(-4,8),\qquad C=(3,7),

and $\angle ABC$ is a right angle. If the radius is $r$ , what is the value of $r^2$ ?

Your answer

Calculator loads as you approach

Find the diameter first, then use only its two ends.

Finish the lesson

5 practice examples left

Finish the remaining questions correctly to complete this lesson.

Quick recap

Name each point once, then reuse the names in distance and midpoint.
midpoint(A,B) gives a point, and distance(A,B) gives a length.
When $A$ and $B$ are the ends of a diameter, their midpoint is the center and half their distance is the radius.
For the ends of a diameter, $r^2$ is a quarter of the squared distance.
From the center to a point on the circle is already one radius, so don't halve it.
A right angle inscribed in a circle makes the side across from it a diameter.
Two points on a circle aren't a diameter unless something tells you so.
Before you submit, reread whether the question wants a radius, $r^2$ , an equation, an area coefficient, a coordinate, or an expression.
Check with a midpoint, equal distances, or a point plugged into your circle equation.

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