# Circles on the SAT

Canonical: https://duckyhelper.com/learn/sat-math/circles/
Updated: 2026-10-01

SAT circle questions use a few tools: the equation \((x - h)^2 + (y - k)^2 = r^2\), which gives the center \((h, k)\) and radius r; arc length and sector area as a fraction of the whole circle; radians; and the fact that a tangent line is perpendicular to the radius. If the equation is expanded, complete the square on x and on y to find the center and radius.

## The key ideas

$$
(x - h)^2 + (y - k)^2 = r^2 \quad\Longrightarrow\quad \text{center } (h, k), \text{ radius } r
$$

The signs flip: \((x + 2)^2\) means \(h = -2\). And the right side is the radius squared, so \(= 49\) means a radius of 7.

An arc or a sector is a fraction of the whole circle. That fraction is the central angle over the full turn:

$$
\text{arc length} = \frac{\theta}{360} \cdot 2\pi r \qquad \text{sector area} = \frac{\theta}{360} \cdot \pi r^2
$$

In radians the full turn is \(2\pi\), and the formulas become arc length \(= r\theta\) and sector area \(= \frac{1}{2}r^2\theta\).

## Worked examples

**Example 1: read the equation**

Problem: What are the center and radius of \((x + 2)^2 + (y - 5)^2 = 49\)?

1. Rewrite \(x + 2\) as \(x - (-2)\), so \(h = -2\) and \(k = 5\).
2. The right side is \(r^2\).

   $$
   r = \sqrt{49} = 7
   $$

Answer: Center \((-2, 5)\), radius 7

**Example 2: complete the square**

Problem: Find the center and radius of \(x^2 + y^2 - 8x + 6y - 11 = 0\).

1. Group the x terms and the y terms, and move the number to the right.

   $$
   (x^2 - 8x) + (y^2 + 6y) = 11
   $$
2. Half of \(-8\) is \(-4\), and \((-4)^2 = 16\). Half of 6 is 3, and \(3^2 = 9\). Add both to each side.

   $$
   (x^2 - 8x + 16) + (y^2 + 6y + 9) = 11 + 16 + 9
   $$
3. Factor each group.

   $$
   (x - 4)^2 + (y + 3)^2 = 36
   $$

Answer: Center \((4, -3)\), radius 6

**Example 3: arc length and sector area**

Problem: A circle has radius 9. A central angle of \(80^\circ\) cuts off an arc. Find the arc length and the sector area.

1. The sector is \(\frac{80}{360} = \frac{2}{9}\) of the circle. Arc length is that fraction of the circumference \(18\pi\).

   $$
   \frac{2}{9} \cdot 18\pi = 4\pi
   $$
2. Sector area is the same fraction of the area \(81\pi\).

   $$
   \frac{2}{9} \cdot 81\pi = 18\pi
   $$

Answer: Arc length \(4\pi\), sector area \(18\pi\)

**Example 4 (SAT-hard): radians, working backward**

Problem: A sector of a circle has a central angle of \(\frac{2\pi}{3}\) radians and an area of \(24\pi\). What is the radius?

1. Use sector area \(= \frac{1}{2}r^2\theta\).

   $$
   \frac{1}{2}r^2 \cdot \frac{2\pi}{3} = 24\pi
   $$
2. Simplify the left side.

   $$
   \frac{\pi r^2}{3} = 24\pi
   $$
3. Multiply by 3 and divide by \(\pi\).

   $$
   r^2 = 72
   $$
4. Take the positive root and simplify: \(\sqrt{72} = \sqrt{36 \cdot 2}\).

Answer: \(r = 6\sqrt{2}\), about 8.49

## Common mistakes

- **Reading \(r^2\) as r.** In \((x - 1)^2 + y^2 = 25\), the radius is 5, not 25. Fix: take the square root of the right side.
- **Keeping the signs inside the parentheses.** \((x + 2)^2\) gives an x-coordinate of \(-2\). Fix: the center has the opposite signs.
- **Adding the completed squares to one side only.** Adding 16 and 9 on the left means adding them on the right too. Fix: balance every step.
- **Using 180 instead of 360.** An \(80^\circ\) arc is \(\frac{80}{360}\) of the circle. Fix: in degrees, the whole circle is 360.
- **Mixing degrees and radians.** In \(\frac{1}{2}r^2\theta\), \(\theta\) must be in radians. Fix: convert before you substitute.

## Quick method

> **Tip: Desmos draws circles**
>
> Type the equation exactly as given, even the expanded form, and Desmos draws the circle. You can read the center and radius from the grid, or drop the answer-choice points in to see which is the center. Completing the square is still worth knowing for questions with unknown constants.

## Practice

**5 SAT-style questions**

1. Which equation is a circle with center \((3, -1)\) and radius 4?
   A. \((x - 3)^2 + (y + 1)^2 = 16\)
   B. \((x + 3)^2 + (y - 1)^2 = 16\)
   C. \((x - 3)^2 + (y + 1)^2 = 4\)
   D. \((x - 3)^2 + (y - 1)^2 = 16\)

   Answer: \((x - 3)^2 + (y + 1)^2 = 16\). Center \((3, -1)\) gives \((x - 3)\) and \((y + 1)\), and the right side is \(4^2 = 16\). The second choice flips both signs, the third forgets to square 4, and the last has the wrong y sign.

2. What is the radius of the circle \(x^2 + y^2 + 10x - 4y = 7\)?
   A. \(\sqrt{7}\)
   B. \(6\)
   C. \(7\)
   D. \(36\)

   Answer: \(6\). Complete the square: add 25 and 4 to both sides to get \((x + 5)^2 + (y - 2)^2 = 36\). The radius is \(\sqrt{36} = 6\). 36 is \(r^2\), and \(\sqrt{7}\) skips completing the square.

3. A circle has radius 12. What is the length of the arc cut off by a central angle of \(45^\circ\)?
   A. \(1.5\pi\)
   B. \(3\pi\)
   C. \(6\pi\)
   D. \(18\pi\)

   Answer: \(3\pi\). \(\frac{45}{360} = \frac{1}{8}\) of the circumference \(24\pi\) is \(3\pi\). \(18\pi\) is the sector area, and \(6\pi\) uses 180 instead of 360.

4. Which point lies inside the circle \((x - 1)^2 + y^2 = 25\)?
   A. \((6, 0)\)
   B. \((4, 3)\)
   C. \((-5, 0)\)
   D. \((1, 5)\)

   Answer: \((4, 3)\). Put each point in the left side and compare with 25. \((4, 3)\): \(9 + 9 = 18 < 25\), inside. \((6, 0)\) and \((1, 5)\) give exactly 25, so they are on the circle, and \((-5, 0)\) gives 36, outside.

5. Student-produced response: a circle centered at the origin passes through the point \((5, 12)\). Its area is \(k\pi\). What is k?

   Answer: 169. The radius is the distance from \((0, 0)\) to \((5, 12)\): \(\sqrt{25 + 144} = 13\). The area is \(\pi (13)^2 = 169\pi\), so \(k = 169\).

## Frequently asked questions

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

Complete the square. Group the x terms and y terms, move the constant to the right, add half of each linear coefficient squared to both sides, and factor. Then read \((h, k)\) and take the square root of the right side for r.

### What is the difference between arc length and sector area?

Arc length is the distance along the curved edge, a length. Sector area is the space inside the pie slice, an area. Both are the same fraction of the whole circle: of the circumference for arc length, and of the area for sector area.

### What does it mean that a tangent is perpendicular to the radius?

A tangent line touches the circle at one point. The radius drawn to that point meets the tangent at a right angle. SAT questions use that right angle to build a right triangle and apply the Pythagorean theorem.

## Sources

- [College Board: SAT Math, Geometry and Trigonometry skills (circles)](https://satsuite.collegeboard.org/sat/whats-on-the-test/math/types/geometry-trigonometry), accessed 2026-10-01

## Related

- [Completing the square](https://duckyhelper.com/learn/algebra-2/completing-the-square/)
- [Right triangles and trigonometry on the SAT](https://duckyhelper.com/learn/sat-math/right-triangles-and-trigonometry/)
- [Circle theorems: angles, chords, tangents and arcs](https://duckyhelper.com/learn/geometry/circle-theorems/)
- [Area and volume on the SAT](https://duckyhelper.com/learn/sat-math/area-and-volume/)
- [SAT Math study guides](https://duckyhelper.com/learn/sat-math/)

## Try asking Ducky

- "Why is the center (-2, 5) and not (2, -5)?"
- "Walk me through completing the square on this circle."
- "How do I know if I should use degrees or radians here?"

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