Circles

Circles show up across multiple question types on the Digital SAT, covering everything from arc length and sector area to circle equations in the coordinate plane. You can expect roughly 2–4 questions on circles per test, spanning both the easier and harder modules. This topic rewards you for knowing a handful of formulas cold and for being comfortable with algebraic techniques like completing the square. Since a calculator is available on every math question, the computational side is manageable, but you still need to set problems up correctly.

Circle Basics: Radii, Diameter, and Key Properties

Every circle is defined by its center and its radius. The radius is the distance from the center to any point on the circle. The diameter is twice the radius and passes through the center, connecting two points on the circle. All radii of a given circle are equal in length.

A tangent line touches a circle at exactly one point. The critical property: a tangent is always perpendicular to the radius drawn to the point of tangency. This 90° angle frequently creates right triangles in SAT problems, which you can then solve with the Pythagorean theorem.

Example: A circle has center $O$ and radius 5. A tangent line from external point $P$ touches the circle at point $T$. If $OP = 13$, what is the length of $PT$?

Since $OT \perp PT$, triangle $OTP$ is a right triangle with hypotenuse $OP = 13$ and leg $OT = 5$.

$PT = \sqrt{OP^2 - OT^2} = \sqrt{169 - 25} = \sqrt{144} = 12$

Arc Length and Sector Area

An arc is a portion of a circle's circumference. A sector is the "pie slice" region between two radii and the arc they intercept. Both depend on what fraction of the full circle the central angle represents.

Using degrees:

$\text{arc length} = \frac{\theta}{360} \times 2\pi r$

$\text{sector area} = \frac{\theta}{360} \times \pi r^2$

Using radians:

$\text{arc length} = r\theta$

$\text{sector area} = \frac{1}{2}r^2\theta$

The radian formulas are cleaner, which is why radians matter.

Example 1: A circle has radius 9. An arc subtends a central angle of 60°. What is the arc length?

$\text{arc length} = \frac{60}{360} \times 2\pi(9) = \frac{1}{6} \times 18\pi = 3\pi$

Example 2: A circle has radius 10. A sector has a central angle of $\frac{3\pi}{5}$ radians. What is the sector area?

$\text{sector area} = \frac{1}{2}(10)^2\left(\frac{3\pi}{5}\right) = \frac{1}{2}(100)\left(\frac{3\pi}{5}\right) = 30\pi$

Converting Between Degrees and Radians

The SAT expects you to move fluently between degrees and radians. The key relationship:

$180° = \pi \text{ radians}$

To convert degrees to radians, multiply by $\frac{\pi}{180}$.

To convert radians to degrees, multiply by $\frac{180}{\pi}$.

Example: Convert 135° to radians.

$135 \times \frac{\pi}{180} = \frac{135\pi}{180} = \frac{3\pi}{4}$

Example: Convert $\frac{5\pi}{6}$ radians to degrees.

$\frac{5\pi}{6} \times \frac{180}{\pi} = \frac{5 \times 180}{6} = 150°$

Common conversions worth memorizing: 30° = $\frac{\pi}{6}$, 45° = $\frac{\pi}{4}$, 60° = $\frac{\pi}{3}$, 90° = $\frac{\pi}{2}$, 180° = $\pi$, 360° = $2\pi$.

The Unit Circle and Trigonometric Ratios

The unit circle is a circle centered at the origin with radius 1. Any point on the unit circle can be written as $(\cos\theta, \sin\theta)$, where $\theta$ is the angle measured counterclockwise from the positive $x$-axis.

This means that for standard angles on the unit circle:

The SAT may ask you to evaluate trig functions at these angles or identify which angle produces a given sine or cosine value. Knowing the unit circle also helps you reason about angles in other quadrants, where the signs of sine and cosine change based on the signs of $x$ and $y$.

Example: In the unit circle, what is the value of $\sin\left(\frac{5\pi}{6}\right)$?

The angle $\frac{5\pi}{6}$ is in the second quadrant (between $\frac{\pi}{2}$ and $\pi$). Its reference angle is $\pi - \frac{5\pi}{6} = \frac{\pi}{6}$. Sine is positive in the second quadrant, so:

$\sin\left(\frac{5\pi}{6}\right) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$

Circle Equations in the Coordinate Plane

The standard form equation of a circle with center $(h, k)$ and radius $r$ is:

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

Every ordered pair $(x, y)$ that satisfies this equation lies on the circle. The equation works because it's really the distance formula in disguise: the distance from any point $(x, y)$ on the circle to the center $(h, k)$ equals $r$.

Reading the Equation

Watch the signs carefully. The standard form uses subtraction, so:

• $(x - 3)^2 + (y + 2)^2 = 25$ has center $(3, -2)$ and radius $\sqrt{25} = 5$
• $(x + 1)^2 + (y - 4)^2 = 7$ has center $(-1, 4)$ and radius $\sqrt{7}$

A very common trap: confusing the signs of $h$ and $k$. If you see $(y + 2)^2$, that means $k = -2$, not $+2$.

How Changes Affect the Graph

• Replacing $h$ with a larger value shifts the circle right. Replacing $h$ with a smaller value shifts it left.
• Replacing $k$ with a larger value shifts the circle up. Replacing $k$ with a smaller value shifts it down.
• Increasing $r^2$ on the right side makes the circle larger. Decreasing it makes the circle smaller.

Example: A circle has equation $(x - 2)^2 + (y + 5)^2 = 16$. If the circle is shifted 3 units to the left and 1 unit up, what is the new equation?

The original center is $(2, -5)$. Shifting left 3: $2 - 3 = -1$. Shifting up 1: $-5 + 1 = -4$. The radius doesn't change.

$(x + 1)^2 + (y + 4)^2 = 16$

Completing the Square for Circle Equations

Many SAT circle equations appear in expanded (general) form:

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

To find the center and radius, you need to rewrite this in standard form by completing the square.

Step-by-step process:

1. Group $x$-terms and $y$-terms together. Move the constant to the other side.
2. For each group, take half the linear coefficient, square it, and add it to both sides.
3. Factor each group into a perfect square binomial.
4. Read off center and radius.

Example: The equation $x^2 + y^2 + 8x - 6y = 0$ represents a circle. What is the radius?

Group and move the constant (already 0 on the right):

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

Complete the square for $x$: half of 8 is 4, and $4^2 = 16$. Add 16 to both sides.

Complete the square for $y$: half of $-6$ is $-3$, and $(-3)^2 = 9$. Add 9 to both sides.

$(x^2 + 8x + 16) + (y^2 - 6y + 9) = 0 + 16 + 9$

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

The center is $(-4, 3)$ and the radius is $\sqrt{25} = 5$.

Example 2: Find the center and radius of $x^2 + y^2 - 10x + 2y + 17 = 0$.

$(x^2 - 10x) + (y^2 + 2y) = -17$

Half of $-10$ is $-5$, squared is 25. Half of 2 is 1, squared is 1.

$(x^2 - 10x + 25) + (y^2 + 2y + 1) = -17 + 25 + 1$

$(x - 5)^2 + (y + 1)^2 = 9$

Center: $(5, -1)$, radius: $3$.

Using the Distance Formula with Circles

Since a circle is the set of all points at distance $r$ from the center, the distance formula connects directly to circles. If you're given two endpoints of a diameter, you can find the center (the midpoint) and the radius (half the distance between the endpoints), then write the equation.

Example: A diameter of a circle has endpoints $(1, 2)$ and $(7, 10)$. Write the equation of the circle.

Center (midpoint): $\left(\frac{1+7}{2}, \frac{2+10}{2}\right) = (4, 6)$

Diameter length: $\sqrt{(7-1)^2 + (10-2)^2} = \sqrt{36 + 64} = \sqrt{100} = 10$

Radius: $\frac{10}{2} = 5$

Equation: $(x - 4)^2 + (y - 6)^2 = 25$

What to Watch For on Test Day

1. Sign errors in circle equations. When you see $(x + 3)^2$, the center's $x$-coordinate is $-3$, not $+3$. This is the single most common mistake on these questions.

2. $r$ vs. $r^2$. The right side of the standard equation is $r^2$. If the equation says $= 49$, the radius is 7, not 49. Conversely, if a problem gives you radius 5, you write $= 25$.

3. Completing the square: add to BOTH sides. When you add a number to complete the square on the left, you must add the same number to the right. Forgetting this gives you the wrong radius every time.

4. Arc and sector formulas: match your angle units. If the angle is in degrees, use the fraction $\frac{\theta}{360}$. If it's in radians, use $r\theta$ for arc length or $\frac{1}{2}r^2\theta$ for sector area. Mixing these up produces wrong answers that often match trap choices.

5. Tangent lines create right angles. Whenever a tangent meets a radius, you have a 90° angle. Look for right triangles you can solve with the Pythagorean theorem or trig ratios. This setup appears frequently and is easy points once you spot it.