SAT Circle Equations: Find the Center and Radius Without Sign Mistakes

The standard equation of a circle is:

$$\boxed{(x-h)^2+(y-k)^2=r^2}$$

where:

• center: $\boxed{(h,k)}$
• radius: $\boxed{r}.$

The most common SAT mistake is reading the signs inside the parentheses directly.

If:

$$(x+3)^2+(y-5)^2=49,$$

then:

$$x+3=x-(-3),$$

so the center is:

$$\boxed{(-3,5)}$$

and the radius is:

$$\boxed{7}.$$

The sign inside each square is opposite the center coordinate.

Why circle equations matter on the SAT

College Board's Geometry and Trigonometry domain explicitly includes circles. Questions can require you to:

• identify a center;
• identify a radius;
• write a circle equation;
• recognize whether a point lies on a circle;
• rewrite an expanded equation by completing the square;
• connect a circle to tangent or coordinate geometry facts.

For the broader topic, see SAT Circles. If the equation is expanded, review Completing the Square.

Read standard form quickly

Given:

$$(x-4)^2+(y+2)^2=36,$$

center:

$$\boxed{(4,-2)}$$

radius:

$$\sqrt{36}=6.$$ $$\boxed{6}$$

The right side is $r^2$, not $r$.

Worked example 1: sign trap

$$(x+7)^2+(y-1)^2=25.$$

Center:

$$\boxed{(-7,1)}.$$

Radius:

$$\boxed{5}.$$

A distractor may offer $(7,-1)$ by copying the visible signs.

Equation from center and radius

Suppose center:

$$(2,-5)$$

and radius:

$$4.$$

Substitute:

$$(x-2)^2+(y-(-5))^2=4^2.$$ $$\boxed{(x-2)^2+(y+5)^2=16}.$$

Do not forget to square the radius.

Worked example 2: center on an axis

Center:

$$(0,6)$$

radius:

$$3.$$

Equation:

$$(x-0)^2+(y-6)^2=9.$$

So:

$$\boxed{x^2+(y-6)^2=9}.$$

If the center is at the origin:

$$\boxed{x^2+y^2=r^2}.$$

Point on a circle

A point $(x,y)$ lies on the circle if it satisfies the equation exactly.

Example:

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

Test point:

$$(5,3).$$ $$(5-2)^2+(3+1)^2 = 9+16 = 25.$$

So:

$$\boxed{(5,3)\text{ lies on the circle}}.$$

Inside, on, or outside

For circle:

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

compute the left side for a point.

• left side $<r^2$ → inside;
• left side $=r^2$ → on;
• left side $>r^2$ → outside.

This is just comparing squared distance from the center with squared radius.

Worked example 3: point classification

Circle:

$$x^2+y^2=25.$$

Point:

$$(3,2).$$ $$3^2+2^2=13<25.$$

So:

inside

Point:

$$(4,3)$$

gives:

$$16+9=25.$$

So it is on the circle.

Circle equation is a distance equation

The distance from $(x,y)$ to center $(h,k)$ is:

$$\sqrt{(x-h)^2+(y-k)^2}.$$

Setting that distance equal to $r$:

$$\sqrt{(x-h)^2+(y-k)^2}=r.$$

Squaring gives:

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

This explains the standard form rather than making it a memorized template.

Expanded circle equations

The SAT may give:

$$x^2+y^2-6x+8y=0.$$

The center and radius are hidden.

Group x and y terms:

$$(x^2-6x)+(y^2+8y)=0.$$

Complete each square.

For x:

$$x^2-6x+9=(x-3)^2.$$

For y:

$$y^2+8y+16=(y+4)^2.$$

Add 9 and 16 to both sides:

$$(x-3)^2+(y+4)^2=25.$$

Therefore:

$$\boxed{\text{center }(3,-4)}$$

and:

$$\boxed{r=5}.$$

Worked example 4: constant on the left

Rewrite:

$$x^2+y^2+4x-10y-7=0.$$

Move constant:

$$x^2+4x+y^2-10y=7.$$

Complete x-square:

$$x^2+4x+4=(x+2)^2.$$

Complete y-square:

$$y^2-10y+25=(y-5)^2.$$

Add 4 and 25 to the right:

$$(x+2)^2+(y-5)^2=36.$$

Center:

$$\boxed{(-2,5)}$$

radius:

$$\boxed{6}.$$

Fast center shortcut from expanded form

For a circle with equal x² and y² coefficients and no xy term:

$$x^2+y^2+Dx+Ey+F=0,$$

the center is:

$$\left(-\frac D2,-\frac E2\right).$$

Example:

$$x^2+y^2-8x+6y+1=0.$$

Center:

$$\left(4,-3\right).$$

This shortcut is useful, but completing the square is safer when you also need the radius.

Worked example 5: find radius after shortcut

$$x^2+y^2-8x+6y+1=0.$$

Complete squares:

$$(x-4)^2-16+(y+3)^2-9+1=0.$$ $$(x-4)^2+(y+3)^2=24.$$

Radius:

$$\boxed{\sqrt{24}=2\sqrt6}.$$

Do not force the radius to be an integer.

Circle tangent to an axis

If a circle with center $(h,k)$ is tangent to the x-axis, radius equals vertical distance from center to x-axis:

$$\boxed{r=|k|}.$$

If tangent to y-axis:

$$\boxed{r=|h|}.$$

Worked example 6: tangent to x-axis

Center:

$$(4,-7)$$

and circle tangent to x-axis.

Radius:

$$|-7|=7.$$

Equation:

$$\boxed{(x-4)^2+(y+7)^2=49}.$$

Tangent point geometry

A radius to a tangent point is perpendicular to the tangent line.

If a circle is tangent to a horizontal line, the radius to that tangent point is vertical.

If tangent to a vertical line, the radius is horizontal.

This can locate a tangent point quickly.

Circle through a known point

Center:

$$(1,2).$$

Circle passes through:

$$(5,5).$$

Radius squared:

$$(5-1)^2+(5-2)^2 = 16+9 = 25.$$

So:

$$\boxed{(x-1)^2+(y-2)^2=25}.$$

You do not need to calculate $r=5$ separately unless asked.

Worked example 7: find parameter from a point

Circle:

$$(x-3)^2+(y+2)^2=k$$

passes through:

$$(6,2).$$

Substitute:

$$(6-3)^2+(2+2)^2 = 9+16 = 25.$$ $$\boxed{k=25}.$$

Note that $k=r^2$, not radius itself.

Diameter and radius

If endpoints of a diameter are given, center is the midpoint.

Suppose diameter endpoints:

$$(-2,4),\quad(6,10).$$

Center:

$$\left(\frac{-2+6}{2},\frac{4+10}{2}\right) = (2,7).$$

Diameter length:

$$\sqrt{8^2+6^2} = 10.$$

Radius:

$$5.$$

Equation:

$$\boxed{(x-2)^2+(y-7)^2=25}.$$

Common traps

Trap 1: copy signs from parentheses

$$(x+3)^2$$

means center x-coordinate $-3$.

Trap 2: right side is radius

It is radius squared.

Trap 3: add completing-square values only to one side

Maintain equation balance.

Trap 4: forget that radius is positive

Use the positive square root.

Trap 5: use diameter as radius

Divide diameter by 2.

Trap 6: confuse center with tangent point

They are different points connected by a radius.

Trap 7: assume every x²+y²+… equation is a real circle

After completing the square, a negative $r^2$ would not describe a real circle.

A reliable SAT circle-equation routine

1. Look for: $(x-h)^2+(y-k)^2=r^2.$
2. Reverse the inside signs for center.
3. Square-root the right side for radius.
4. If expanded, group x and y terms.
5. Complete both squares.
6. Add balancing constants to the other side.
7. Check whether the question needs center, radius, point membership, or a full equation.

SAT-style practice

Question 1

$$(x-5)^2+(y+4)^2=81.$$

Center:

$$\boxed{(5,-4)}$$

Radius:

$$\boxed{9}$$

Question 2

Center $(-1,6)$, radius 4.

$$\boxed{(x+1)^2+(y-6)^2=16}$$

Question 3

$$x^2+y^2-4x+12y=0.$$

Complete:

$$(x-2)^2+(y+6)^2=40.$$

Center:

$$\boxed{(2,-6)}$$

Radius:

$$\boxed{2\sqrt{10}}$$

Question 4

Circle center $(3,8)$ tangent to x-axis.

Radius:

$$\boxed{8}$$

Question 5

$$(x+2)^2+(y-1)^2=13$$

Does $(1,3)$ lie on circle?

$$3^2+2^2=13.$$

Yes

FAQ: SAT circle equations

What is the standard circle equation?

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

How do I read the center?

Reverse the signs inside the parentheses.

How do I find the radius?

Take the positive square root of the right side.

How do I find center and radius from an expanded equation?

Complete the square separately for x and y.

How do I know if a point lies on a circle?

Substitute the coordinates and check whether the equation is satisfied.

If a circle is tangent to the x-axis, what is the radius?

The absolute value of the center's y-coordinate.

If diameter endpoints are given, how do I find the equation?

Find midpoint for center and half the endpoint distance for radius.

How to practice this on MastaPrep

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See how MastaPrep works or try free SAT practice questions. The free plan needs no credit card.

Related MastaPrep guides

• SAT Circles
• Completing the Square
• SAT Math guides
• SAT Math Formulas

Final takeaway

For circle equations, make two habits automatic:

$$\boxed{(x-h)^2+(y-k)^2=r^2}$$

and:

inside signs reverse for the center

Then remember that expanded circle equations are just standard form hidden behind completing the square.

A circle behind a common factor

Everything above assumes $x^2$ and $y^2$ each have a coefficient of 1. The SAT sometimes hides a circle behind a common factor instead:

$$2x^2+2y^2-8x+12y-6=0.$$

The first instinct is to start completing the square right away, and the 2 in front of each squared term makes a mess of it. Divide every term by 2 before you do anything else:

$$x^2+y^2-4x+6y-3=0.$$

Now it looks like the equations you already know. Group the x-terms and y-terms and move the constant across:

$$(x^2-4x)+(y^2+6y)=3.$$

Half of $-4$ is $-2$, which squares to 4; half of 6 is 3, which squares to 9. Add both to each side:

$$(x-2)^2+(y+3)^2=3+4+9=16.$$

So the center is

$$\boxed{(2,-3)}$$

and the radius is

$$\boxed{4}$$

It is worth one quick check, because the division is where things go wrong. The point $(6,-3)$ is 4 units to the right of the center, so it should sit on the circle. In the original equation: $2(36)+2(9)-8(6)+12(-3)-6=72+18-48-36-6=0$. It does.

The same caution applies if you like the center shortcut $\left(-\frac D2,-\frac E2\right)$: it only works on the divided form. Plug in the undivided coefficients $-8$ and $12$ and you get $(4,-6)$, which is wrong.

Once you have the radius, the question may carry on into an arc or a sector, since the radius feeds straight into circumference $2\pi r$ and area $\pi r^2$. SAT Arc Length vs Sector Area picks up from there.

Before you try the next one

Question 6

A circle in the xy-plane has equation $3x^2+3y^2+18x-12y=0$. What is the radius of the circle?

A) $\sqrt{13}$
B) $13$
C) $\sqrt{39}$
D) $39$

Answer: A. Divide by 3 to get $x^2+y^2+6x-4y=0$, then complete the squares: $(x+3)^2+(y-2)^2=9+4=13$. The right side is $r^2$, so $r=\sqrt{13}$. B stops at $r^2$ and calls it the radius. C and D both come from skipping the division by 3, which turns the right side into $3(9)+3(4)=39$.

Question 7

A circle has center $(-2,3)$ and passes through the point $(1,7)$. Which of the following points also lies on the circle?

A) $(3,8)$
B) $(-5,-1)$
C) $(1,3)$
D) $(5,1)$

Answer: B. Start with $r^2$: $(1+2)^2+(7-3)^2=9+16=25$, so the equation is $(x+2)^2+(y-3)^2=25$. Testing B gives $(-5+2)^2+(-1-3)^2=9+16=25$, so it lies on the circle. D is the trap. It satisfies $(x-2)^2+(y+3)^2=25$, the equation you get by copying the center's signs straight into the parentheses. In the correct equation, D gives $49+4=53$. A gives 50 and C gives 9, so neither of those is on the circle either.

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Official SAT references:
https://satsuite.collegeboard.org/sat/whats-on-the-test/math/types/geometry-trigonometry
https://satsuite.collegeboard.org/practice/student-question-bank/math

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