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SAT math circle equations: find the center and radius
Learn how a circle’s equation describes its center and radius, how to graph it, and how to rewrite expanded equations by completing the square.
SPSAT Practice Editorial8 min read
The short version
What you'll take away
- A circle is the set of points a fixed distance, called the radius, from its center.
- In (x − h)² + (y − k)² = r², the center is (h, k) and the radius is r.
- Complete the square in x and y to convert an expanded equation into standard form.
In this article
A circle keeps the same distance from its center
A circle is made of all points that are exactly the same distance from one fixed point. That fixed point is the center, and the distance is the radius. On a coordinate plane, the distance formula lets us describe every point (x, y) on the circle.
If the center is (h, k) and the radius is r, every point on the circle satisfies \(\displaystyle (x-h)^2 + (y-k)^2 = r^2\). The squared horizontal and vertical distances add to the squared radius, just as the legs of a right triangle add to the hypotenuse squared.
The study lab
Center and radius describe this circle
Center (2, -1), radius 3. The purple segment runs from the center to a point on the circle.
Read the center and radius from standard form
The standard equation is \(\displaystyle (x-h)^2 + (y-k)^2 = r^2\). The center coordinates are h and k, and the radius is the positive square root of the right side. If the right side is 25, the radius is 5, not 25.
For example, \(\displaystyle (x-2)^2 + (y+1)^2 = 9\) has center (2, -1) and radius 3. Notice the signs inside the parentheses: \(\displaystyle y+1\) is the same as \(\displaystyle y-(-1)\), so the center’s y-coordinate is -1.
- The x-part \(\displaystyle (x-h)^2\) gives the center’s x-coordinate h.
- The y-part \(\displaystyle (y-k)^2\) gives the center’s y-coordinate k.
- Take the square root of the right side to get the radius.
Write an equation from a center and radius
Substitute the center and radius directly into standard form. For a circle centered at (-3, 2) with radius 4, use h = -3, k = 2, and r = 4: \(\displaystyle (x-(-3))^2 + (y-2)^2 = 4^2\). Simplify to \(\displaystyle (x+3)^2 + (y-2)^2 = 16\).
You can check the equation with a point one radius to the right of the center. That point is (1, 2). Substitution gives \(\displaystyle (1+3)^2 + (2-2)^2 = 16\), so it lies on the circle.
Convert expanded form by completing the square
An equation may be expanded instead of written in standard form. Consider \(\displaystyle x^2 + y^2 - 4x + 6y - 12 = 0\). First move the constant: \(\displaystyle x^2 - 4x + y^2 + 6y = 12\). Group the x-terms and y-terms.
To complete a square, take half the linear coefficient and square it. Half of -4 is -2, and \(\displaystyle (-2)^2=4\). Half of 6 is 3, and \(\displaystyle 3^2=9\). Add 4 and 9 to both sides so the equation stays balanced:
- \(\displaystyle (x^2 - 4x + 4) + (y^2 + 6y + 9) = 12 + 4 + 9\).
- \(\displaystyle (x-2)^2 + (y+3)^2 = 25\).
- The center is (2, -3), and the radius is 5.
Check whether a point is on the circle
Substitute the point’s x- and y-coordinates into the equation. If the left side equals the right side, the point is on the circle. If it is smaller or larger, the point is inside or outside, respectively.
For \(\displaystyle (x-2)^2 + (y+1)^2 = 9\), test (5, -1): \(\displaystyle (5-2)^2 + (-1+1)^2 = 9 + 0 = 9\). The point is on the circle. The center itself gives 0 on the left, so it is inside.
Common mistakes to avoid
The center’s signs are opposite the signs that appear inside the squared parentheses. The radius is the square root of the right side. When completing the square, add the same values to both sides.
A circle centered at the origin has the simpler equation \(\displaystyle x^2+y^2=r^2\). Do not confuse the radius with the diameter: the diameter is twice the radius.
- Center (−4, 3) means the squared terms are (x + 4)² and (y − 3)².
- If the equation ends in 49, the radius is 7.
- Use parentheses when substituting negative coordinates.
Practice: try these before checking the answers
For each question, identify the form first. If it is expanded, collect x-terms and y-terms before completing the square.
- 1. Find the center and radius of \(\displaystyle (x+3)^2 + (y-2)^2 = 16\).
- 2. Write the equation of a circle centered at (2, -1) with radius 3.
- 3. Is (5, -1) on the circle \(\displaystyle (x-2)^2+(y+1)^2=9\)?
- 4. Rewrite \(\displaystyle x^2+y^2-4x+6y-12=0\) in standard form and identify its center and radius.
- 5. A circle centered at the origin passes through (0, -6). Write its equation.
A quick circle-equation check for SAT Math
Read the standard form carefully: reverse the signs in the parentheses to locate the center, and take the square root of the constant to find the radius. If the equation is expanded, complete the square in both variables and keep both sides balanced.
Then check your result with a point you know is on the circle, such as a point one radius to the right, left, up, or down from the center. A quick substitution can catch a sign error before you move on.
Sources & further reading
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