SAT Circle Equations: Find the Center and Radius

To find a circle’s center and radius, rewrite its equation as $(x-h)^2+(y-k)^2=r^2$. The center is $(h,k)$, and the radius is the positive square root of the right side. For an expanded equation, first make the coefficients of both squared terms equal to one, then complete the square separately for the $x$-terms and $y$-terms.

College Board’s SAT Math skill descriptions include graphs of circles, the distance formula, and completing the square in a circle equation. The examples here are original Kibo practice, not official SAT questions or a prediction of how often this skill will appear.

What the standard form tells you

A circle consists of points a fixed positive distance from its center. For a point $(x,y)$ and center $(h,k)$, the squared horizontal and vertical distances add to the squared radius:

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

This is the distance formula with the distance fixed at the radius. OpenStax’s guide to circle equations explains standard form and completing the square. For the right-triangle connection, review Kibo’s coordinate-distance worked examples.

For example, consider $(x+7)^2+(y-2)^2=121$. Since $x+7=x-(-7)$, the center is $(-7,2)$. The radius is $\sqrt{121}=11$. The point $(4,2)$ is $11$ units horizontally from that center; substituting gives $(4+7)^2+(2-2)^2=121$.

The center is not $(7,-2)$, and the radius is not $121$. Match each parenthesis to subtraction of a center coordinate, then take the square root of the constant.

Complete the square in an expanded equation

Find the center and radius of $x^2+y^2-10x+14y-95=0$. Start by grouping each variable’s terms and moving the constant:

$$(x^2-10x)+(y^2+14y)=95$$

To make $x^2+bx$ a perfect square, add $(b/2)^2$. For the $x$-group here, half of $-10$ is $-5$, whose square is $25$. For the $y$-group, half of $14$ is $7$, whose square is $49$. Add both amounts to both sides:

$$(x^2-10x+25)+(y^2+14y+49)=95+25+49$$

$$(x-5)^2+(y+7)^2=169$$

The center is $(5,-7)$, and the radius is $\sqrt{169}=13$. Check the rewrite by expanding: the left side becomes $x^2-10x+25+y^2+14y+49$. Subtracting $169$ gives the original constant $25+49-169=-95$.

Adding $25$ and $49$ only on the left would change the circle. Likewise, $(x-10)^2$ would produce a middle term of $-20x$, so it cannot replace $x^2-10x+25$. Halve the linear coefficient before squaring. The same algebra appears in the quadratic vertex guide.

Check whether a point is on the circle

For the circle just found, test $(10,5)$ by substituting both coordinates into standard form:

$$(10-5)^2+(5+7)^2=25+144=169$$

The result equals $r^2$, so the point is on the circle. The original expanded equation confirms it: $100+25-100+70-95=0$. A point must satisfy the equation, not merely share one coordinate with the center.

By comparison, $(10,-2)$ gives $(10-5)^2+(-2+7)^2=25+25=50$. Since $50<169$, that point is inside the circle and does not satisfy its equation. In standard form, a squared distance larger than $r^2$ would place a point outside.

Divide first when both squared terms have a common coefficient

Find the center and radius of $3x^2+3y^2+12x-30y-21=0$. Divide every term by $3$, including the linear terms and constant:

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

Group and complete the squares. Half of $4$ is $2$, and half of $-10$ is $-5$:

$$(x^2+4x+4)+(y^2-10y+25)=7+4+25$$

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

The center is $(-2,5)$, and the radius is $6$. As a check, $(4,5)$ is six units to the right of the center. In the original equation, it gives $48+75+48-150-21=0$.

This method assumes the equation has equal, nonzero coefficients on $x^2$ and $y^2$, with no $xy$ term. Do not force a different equation into the circle formula. After completing the squares, a positive right side gives a circle; zero gives a single point, and a negative value gives no real points.

Try four original practice questions

Keep the radius exact. Show your completed squares when needed, and use the original equation for any point check.

1. Find the center and radius: $(x-8)^2+(y+1)^2=20$.
2. Find the center and radius: $x^2+y^2+6x-16y+24=0$.
3. Find the center and radius of $2x^2+2y^2-28x+8y+56=0$. Is $(10,2)$ on this circle?
4. Write a circle equation with center $(-6,-5)$ passing through $(2,1)$.

Use a short checking routine

Before finishing, check that you divided every term by any common squared-term coefficient, added the square-completion amounts to both sides, read the center signs correctly, and took the square root for the radius. Expanding your final equation should recover the original one.

Record the first step that went wrong in your practice error log, then retry that problem later with the solution hidden. Try a free SAT practice question with Kibo and explain how you checked your answer.

Prepared with AI assistance. The worked examples and practice questions are original; their equations, centers, squared radii, and point checks were verified separately with exact arithmetic. No human review or score improvement is claimed. Sources checked November 10, 2026. Kibo is not affiliated with or endorsed by College Board.