Circle geometry with distance and midpoint

Your feedback matters:

Lesson progressPractice problems 0/5

Difficulty: Advanced
Estimated time: 55 minutes
Domains: Geometry and Trigonometry
Techniques: CirclesDistanceMidpointDiameterCoordinate-geometryInscribed-angles

What you’ll learn

Define coordinate points and use distance and midpoint.
Turn diameter endpoints into a circle’s center, radius, squared radius, and equation.
Use a center and one boundary point to find a radius or area.
Recognize when a right inscribed angle makes the opposite side a diameter.
Avoid premature rounding and verify exact values.
Choose when these commands are useful and when hand reasoning is shorter.

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 coordinate points instead of a ready-made equation. If the points are the endpoints of a diameter, Desmos can find the center and radius in two short commands:

midpoint(A,B) gives the center.
distance(A,B)/2 gives the radius.

That information can unlock the circle’s equation or area.

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?

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
$(x+1)^2+(y-1)^2=10$

Fast Desmos solution

Name the endpoints $A$ and $B$ , then enter midpoint(A,B) and distance(A,B)/2. Desmos returns the center $(-1,1)$ and radius $5$ . Using center-radius form,

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

the circle is

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

The answer is A.

The commands do the arithmetic. You still need to recognize that the given segment is a diameter and report the exact quantity the prompt requests.

Calculator loads as you approach

The midpoint is the center, and half the endpoint distance is the radius.

When should you use this method?

Choose the shortest method that preserves the circle’s exact information.

Use distance or midpoint when…

the problem gives the two endpoints of a diameter;
the problem gives the center and one point on the circle;
an inscribed right triangle has all three vertices on the circle and hides a diameter; or
the coordinate differences are awkward to simplify mentally.

Use hand reasoning first when…

the points share an x- or y-coordinate and their separation is obvious;
the midpoint is immediately visible;
center-radius form already gives the center and radius; or
the problem is mainly about another circle theorem.

Two points on a circle are not automatically diameter endpoints. The prompt or a circle theorem must justify that relationship.

If you know three boundary points and no diameter is identified, use geometry to coordinates to circle regression.

Check your understanding:

A circle has center $(3,-1)$ and passes through $(3,6)$ . Should you begin with the distance command or mental math? What is the radius?

1. Name the points, then use the commands

One point definition prevents repeated typing

Suppose

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

Enter each point on its own line:

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

Then enter:

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

The commands return

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

and

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

The midpoint is a point, so its output has two coordinates. The distance is a length, so its output is one nonnegative number.

The order of the points does not change either result.

Try it yourself:

Change $B$ from $(4,9)$ to $(4,-7)$ . Predict the new midpoint and distance before reading the outputs, then reset the example.

Check your understanding:

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

Common mistake:

Pairing the wrong coordinates. Enter each ordered pair exactly as given. The first coordinate is $x$ and the second is $y$ . Naming the points once avoids rebuilding the pairs differently in the two commands.

Calculator loads as you approach

Name each point once, then reuse A and B in both commands.

2. Convert diameter endpoints into circle information

Midpoint gives the center

If $A=(x_1,y_1)$ and $B=(x_2,y_2)$ are the endpoints of a diameter, the center is

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

In Desmos, enter C=midpoint(A,B).

Distance gives the diameter, not the radius

The distance between opposite endpoints spans the full circle. Therefore,

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

If an equation or area needs $r^2$ , enter (distance(A,B)/2)^2. This avoids manually rounding an irrational radius before squaring it.

For $A=(-3,2)$ and $B=(5,8)$ , the midpoint is $(1,5)$ . The endpoint distance is

\sqrt{8^2+6^2}=10,

so the radius is $5$ 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:

Using the endpoint distance as the radius. The distance between opposite diameter endpoints crosses the entire circle. Divide by $2$ , or divide the squared distance by $4$ when you need $r^2$ .

Calculator loads as you approach

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

3. Use the center and one boundary point

Their distance is one radius

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

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

There is no division by $2$ because $\overline{CP}$ already goes from the center to the boundary.

Worked example

A circular stage has center $C=(2,-1)$ and passes through $P=(8,5)$ . What is its area in terms of $\pi$ ?

The squared radius is

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

Therefore,

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

Desmos displays the radius as a decimal because $r=6\sqrt2$ . Evaluating r^2 displays $72$ without manually rounding. Because the calculator result is numerical, verify the exact value from $6^2+6^2=72$ .

Try it yourself:

Change $P$ to $(5,3)$ . Predict $r^2$ and the area coefficient before reading 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:

Halving a radius. Divide by $2$ only when the two points are opposite endpoints of a diameter. The segment from the center to a point on the circle is already one radius.

Calculator loads as you approach

The center-to-boundary distance is the radius. Squaring it displays 72, which the coordinate calculation verifies exactly.

4. Let a right inscribed angle reveal the diameter

A triangle is inscribed in a circle when all three vertices lie on the circle. If one angle is $90^\circ$ , the side opposite that angle is a diameter.

For triangle $ABC$ , if $\angle ACB=90^\circ$ , then $\overline{AB}$ is a diameter. This turns a theorem problem into the diameter workflow from Section 2.

Check your understanding:

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

Example: Find a circle from an inscribed right angle

Worked example

Triangle $ABC$ is inscribed in a circle in the $xy$ -plane. Its vertices include

$A=(-2,1)$

$B=(6,7)$

and $C=(5,0)$ .

$\angle ACB$ 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

Identify the diameter

Because $\angle ACB$ is a right inscribed angle, the opposite side $\overline{AB}$ is a diameter. Point $C$ confirms the triangle’s location, but $A$ and $B$ are enough to determine the circle.

Step 2

Find the center and radius

Enter $A$ and $B$ , then use midpoint(A,B) and 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

The hypotenuse AB is the diameter, so its midpoint and half-length determine the circle.

Step 3

Write the requested equation

With center $(2,4)$ and radius $5$ ,

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

The answer is A. As a check, $C=(5,0)$ satisfies the equation because

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

Common mistake:

Using a leg as the diameter. The diameter is the side opposite the $90^\circ$ angle, which is the hypotenuse.

5. Keep the result exact and answer the prompt

The calculator output is only an intermediate result. Before submitting:

Identify whether distance represents a radius or a diameter.
Decide whether the prompt asks for $r$ , $r^2$ , an equation, an area, or a coordinate.
Avoid rounding before the final step.
Verify with a midpoint, coordinate calculation, or point substitution.

For diameter endpoints,

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

For a center $(h,k)$ and boundary point $P$ ,

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

If Desmos shows a decimal radius, use its square to identify a candidate value, then verify that value exactly from the coordinates.

Check your understanding:

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

Finish the solution

The endpoints and center calculation are ready. Complete the radius and area work, then submit the coefficient of $\pi$ .

Finish the circle area

Finish the solution

In the $xy$ -plane, the endpoints of a diameter 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), which Desmos shows as $(2,2)$ .
Find the diameter with distance(A,B).
Divide by $2$ to get the radius.
Square the radius to find $k=r^2$ .

Finish it

Your answer

Calculator loads as you approach

Add the diameter, radius, and squared-radius calculations to finish.

Practice problems

SAT practice problems

Decide what the points represent before using a command. Choose the shortest valid method, then verify the exact quantity requested.

Write the circle equation

Practice problem

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

Calculator loads as you approach

Find the midpoint and half the endpoint distance, then write center-radius form.

Find the area coefficient

Practice problem

A circular stage 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

The center-to-boundary distance is the radius. Square it before answering.

Find an area coefficient

Practice problem

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

Your answer

Find squared radius from a hidden diameter

Practice problem

Triangle $ABC$ is inscribed in a circle. Its 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

Identify the diameter first, then use only its two endpoints.

Finish the lesson

5 practice examples left

Finish the remaining questions correctly to complete this lesson.

Quick recap

Name coordinate points once, then reuse them in distance and midpoint.
midpoint(A,B) returns a point; distance(A,B) returns a length.
For diameter endpoints, the midpoint is the center and half the distance is the radius.
For a center and boundary point, their distance is already the radius.
A right inscribed angle makes the opposite side a diameter.
Two arbitrary points on a circle do not identify a diameter.
Verify numerical calculator output exactly from the coordinates.
Reread the prompt before submitting a radius, $r^2$ , equation, area coefficient, or coordinate.

Next lesson

Regression failure modes: exact fit or best fit?

Decide when a fitted model represents exact conditions and when it only approximates data.

Start next lesson

Free Practice

Free question bank practice problems

Circles