Math · Geometry and Trigonometry

SAT Circles: The Complete Guide

Q: What is the standard form of a circle's equation?

(x − h)² + (y − k)² = r², where (h, k) is the center and r is the radius. Read the center by reversing the signs inside the parentheses.

Q: How do I find the center of a circle from its expanded equation?

Group the x terms and the y terms, complete the square in each group, and read the center from the squared binomials. Remember to add the same numbers to both sides.

Q: Is arc length measured in radians or degrees?

Either, as long as the formula matches the unit. Use s = rθ with θ in radians, or use the fraction of 360 degrees with degrees. Do not mix the two in one step.

Q: Is a tangent line perpendicular to the radius?

Yes. At the point of tangency, the radius and the tangent line form a right angle, which is often the key to using the Pythagorean theorem in a tangent problem.

Q: How can I use Desmos on circle questions?

Type the equation in standard form and check whether its graph matches the circle in the problem. Use Desmos to verify your center and radius, not to replace the algebra.

Circle questions ask for circumference and area, arc lengths and sectors, angles in a circle, tangent lines, and the equation of a circle in the coordinate plane. Most points come from converting an angle into a fraction of a full turn and from completing the square correctly. This guide covers the circle rules the SAT uses, shows how to switch between degrees and radians, and gives worked examples from easy to hard.

Section: Math, Geometry and Trigonometry
Calculator: Desmos is useful for checking a circle equation against its graph
Angle units: Degrees or radians; a full turn is 360 degrees or 2π radians
Answer types: Multiple choice and student-produced response

What this skill tests

A circle question asks about a measurement tied to the circle's radius, a central or inscribed angle, a tangent line, or an equation in $x$ and $y$ . Typical stems look like these:

A sector has a given central angle. Find its arc length or its area.
Points lie on a circle. Find an inscribed angle or the arc it intercepts.
A line is tangent to a circle. Find a length or an angle.
A circle is given as an equation in expanded form. Find its center and radius.

The SAT often gives angles in radians on these questions, so be ready to read $\pi$ as part of a fraction of the full turn. A full turn is $2\pi$ radians.

The core idea at a glance

Circle problems reduce to two conversions. First, convert an angle into a fraction of a full circle: $\frac{\theta}{360^\circ}$ when the angle is in degrees, or $\frac{\theta}{2\pi}$ when it is in radians. Second, when the circle is given as an equation, complete the square to reveal the center $(h,k)$ and the radius $r$ .

Circles also come with a few geometric facts the SAT uses often. A radius meets a tangent at a right angle. An inscribed angle is half its intercepted arc. A diameter subtends a right angle at any point on the circle.

Core rules, formulas, and equations

C=2\pi r=\pi d,\qquad A=\pi r^2

Circumference and area of a circle with radius

r

and diameter

d

\text{arc length}=\frac{\theta}{360^\circ}\cdot 2\pi r \qquad\text{or}\qquad s=r\theta\ \ (\theta\text{ in radians})

\text{sector area}=\frac{\theta}{360^\circ}\cdot \pi r^2 \qquad\text{or}\qquad A=\frac12 r^2\theta\ \ (\theta\text{ in radians})

A central angle has the same measure as the arc it intercepts. A $70^\circ$ central angle intercepts a $70^\circ$ arc. An inscribed angle is half of the arc it intercepts, so an inscribed angle that faces a semicircle is always $90^\circ$ .

\text{inscribed angle}=\frac12\cdot\text{intercepted arc}

OT^2+PT^2=OP^2

If segment

PT

is tangent to the circle at

T

, and

O

is the center, then triangle

OTP

has a right angle at

T

Equation of a circle. A circle with center $(h,k)$ and radius $r$ has the equation below. Every point on the circle satisfies it, and no other point does.

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

Read the center carefully. In $(x-3)^2$ , the value of $h$ is $3$ . In $(y+5)^2$ , the value of $k$ is $-5$ . The radius is the positive square root of the right-hand side.

Completing the square. To rewrite $x^2+bx$ as a perfect square, add and subtract $\left(\frac b2\right)^2$ . The identity is

x^2+bx=\left(x+\frac b2\right)^2-\frac{b^2}{4}

Apply this to the $x$ terms and the $y$ terms separately, and add the same numbers to the right-hand side so the equation stays balanced.

Step-by-step method

Identify what is given: a radius, a diameter, a central angle, an arc, or an equation.
For an arc or a sector, write the fraction $\frac{\theta}{360^\circ}$ or $\frac{\theta}{2\pi}$ , then decide which quantity you need.
For an equation, group the $x$ terms and the $y$ terms, and move the constant to the right-hand side.
Complete the square in each group, and add the same numbers to both sides.
Read the center from the squared binomials, reversing the sign of each constant inside the parentheses.
Take the square root of the right-hand side to get the radius. A radius is always positive.
Check the answer. A sector's area must be smaller than the whole circle's area, and an arc must be shorter than the circumference.

Worked examples

These examples move from a sector to an equation to a tangent-line problem with a figure. Work each one before you read the steps.

Worked example

Example 1: Area of a sector (easy)

A circle has a radius of $6$ units. A sector of this circle has a central angle of $90^\circ$ . What is the area of the sector, in square units?

A $9\pi$
B $12\pi$
C $18\pi$
D $36\pi$

Show the solution

The area of the whole circle is $\pi r^2=\pi(6)^2=36\pi$ .
The sector is the fraction $\frac{90}{360}=\frac14$ of the circle.
The sector's area is $\frac14(36\pi)=9\pi$ . The answer is A.
Why not C ( $18\pi$ )? That is half the circle, which is what you get by using $180^\circ$ instead of $90^\circ$ .
Why not B ( $12\pi$ )? That is the circumference, $2\pi(6)$ . It is a length, not an area.
Why not D ( $36\pi$ )? That is the area of the whole circle. The sector is only a quarter of it.

Answer: A

Worked example

Example 2: Radius from an expanded equation (medium, student-produced response)

The equation $x^2+y^2-6x+10y=2$ defines a circle in the $xy$ -plane. What is the radius of the circle?

Show the solution

Group the terms: $(x^2-6x)+(y^2+10y)=2$ .
Complete the square. For $x$ , add $\left(\frac{-6}{2}\right)^2=9$ . For $y$ , add $\left(\frac{10}{2}\right)^2=25$ . Add both numbers to the right-hand side as well: $2+9+25=36$ .
The equation is now $(x-3)^2+(y+5)^2=36$ .
The center is $(3,-5)$ , and $r^2=36$ , so $r=6$ .
Trap check: $36$ is $r^2$ , not the radius. Entering $36$ skips the square root.

Answer: 6

Worked example

Example 3: A tangent segment (hard)

Note: Figure not drawn to scale.

Point $P$ lies outside circle $O$ , which has radius $5$ . The distance from $P$ to the center $O$ is $13$ . A segment from $P$ is tangent to the circle at point $T$ . What is the length of $\overline{PT}$ ?

A $8$
B $12$
C $18$
D $169$

Show the solution

A tangent is perpendicular to the radius at the point of tangency, so triangle $OTP$ has a right angle at $T$ .
The radius gives $OT=5$ , and $OP=13$ is the hypotenuse.
Apply the Pythagorean theorem: $PT^2=OP^2-OT^2=169-25=144$ , so $PT=12$ . The answer is B.
Why not A ( $8$ )? That is $13-5$ . Subtracting the radius treats the three points as if they lie on one line, which they do not here.
Why not C ( $18$ )? That is $13+5$ . Adding the radius places the tangent point on the far side of the center, and that is not possible for a tangent segment.
Why not D ( $169$ )? That is $OP^2$ . It forgets to subtract $OT^2$ and to take the square root.

Answer: B

Common traps

The hardest questions

The hardest circle questions combine a right triangle with a circle, use the arc or chord in a nonstandard way, or ask whether a point lies inside a circle. These four patterns cover most of them.

Pattern 1: an inscribed right triangle. If one side of a triangle is a diameter and the third vertex is on the circle, the angle at that vertex is $90^\circ$ . A triangle with diameter $10$ and leg $6$ has its other leg equal to $\sqrt{10^2-6^2}=8$ .

Pattern 2: arc length in radians. An arc of $\frac{2\pi}{3}$ radians on a circle with radius $9$ has length $9\cdot\frac{2\pi}{3}=6\pi$ . Use $s=r\theta$ directly. Converting to degrees first only adds a step.

Pattern 3: inside or outside. Substitute a point into $(x-h)^2+(y-k)^2$ and compare with $r^2$ . For a circle centered at the origin with radius $3$ , the point $(2,2)$ is inside, because $2^2+2^2=8<9$ .

Pattern 4: radius from two points. The radius is the distance from the center to any point on the circle. A circle centered at $(1,2)$ that passes through $(4,6)$ has radius $\sqrt{3^2+4^2}=5$ .

Timing and practice plan

A sector or arc question should take about a minute, and an equation with completing the square may take two minutes. Keep your work organized, because sign errors in the center are the most common mistake on this skill.

Convert ten angles into fractions of a full circle, in degrees and in radians, until you can do it without writing the denominator first.
Solve five sector problems and five arc problems, writing the fraction before the formula.
Complete the square on ten equations, reading the center and radius after each one.
Solve five tangent problems, drawing the right triangle with the radius before you calculate.
Review each error and name it: diameter used, sign of the center, degrees mixed with radians, or a missed square root.

In Solar, repeat these drills with this skill's practice questions. Move on once you can complete the square and find a sector area without a sign or unit error.

Keep reading — it’s free

Create a free Solar account to unlock the rest of this circles guide, plus adaptive practice on exactly this skill.

Core rules, formulas, and equations
Step-by-step method
Worked examples
Common traps
The hardest questions

Unlock the full guide Already have an account? Log in

Circles: FAQ

What is the standard form of a circle's equation?

(x − h)² + (y − k)² = r², where (h, k) is the center and r is the radius. Read the center by reversing the signs inside the parentheses.

How do I find the center of a circle from its expanded equation?

Group the x terms and the y terms, complete the square in each group, and read the center from the squared binomials. Remember to add the same numbers to both sides.

Is arc length measured in radians or degrees?

Either, as long as the formula matches the unit. Use s = rθ with θ in radians, or use the fraction of 360 degrees with degrees. Do not mix the two in one step.

Is a tangent line perpendicular to the radius?

Yes. At the point of tangency, the radius and the tangent line form a right angle, which is often the key to using the Pythagorean theorem in a tangent problem.

How can I use Desmos on circle questions?

Type the equation in standard form and check whether its graph matches the circle in the problem. Use Desmos to verify your center and radius, not to replace the algebra.