Reference Sheet · SAT Math · Geometry
Circles
Quick lookup for everything circle-related on the SAT. Each card includes a brief worked example. For step-by-step teaching, see the Geometry, Trigonometry & Systems notes.
Basic measurements
Circumference
$$C = 2\pi r = \pi d$$
r = radius
d = diameter (= 2r)
d = diameter (= 2r)
The distance all the way around the circle. Use whichever form matches what you're given — radius or diameter.
EXAMPLE
Find the circumference of a circle with radius 4.
$C = 2\pi(4) = $ $8\pi$
Area
$$A = \pi r^2$$
r = radius
The space enclosed inside the circle. Square the radius first, then multiply by π.
EXAMPLE
Find the area of a circle with diameter 10.
$r = 5$, so $A = \pi(5)^2 = $ $25\pi$
Equation of a circle
Standard form (centered at $(h, k)$ with radius $r$)
$$(x - h)^2 + (y - k)^2 = r^2$$
(h, k) = center of the circle
r = radius
r = radius
Watch the signs. A circle centered at $(3, -2)$ gives $(x - 3)^2 + (y + 2)^2 = r^2$ — the signs FLIP because subtracting a negative becomes addition.
EXAMPLE
Write the equation of a circle centered at $(2, -3)$ with radius $5$.
$(x - 2)^2 + (y + 3)^2 = 25$
Notice the $+3$ on the $y$ term: subtracting $-3$ became $+3$.
Try it yourself
- In the panel on the left of the graph, find the three sliders labeled $h$, $k$, and $r$.
- Click and drag the dot on each slider to change its value. Try setting $h = 3$, $k = -2$, $r = 4$.
- Watch the circle on the right move and resize as you drag.
Notice: $h$ shifts the center left/right, $k$ shifts it up/down, and $r$ controls the size — exactly what the formula $(x-h)^2 + (y-k)^2 = r^2$ predicts.
If you're given expanded form
$$x^2 + y^2 + Dx + Ey + F = 0$$
Complete the square in $x$ and in $y$ separately to convert to standard form. (See the Quadratics notes for the completing-the-square procedure.)
EXAMPLE
Find the center and radius of $x^2 + y^2 - 6x + 4y - 12 = 0$.
Group: $(x^2 - 6x) + (y^2 + 4y) = 12$
For $x$: half of $-6$ is $-3$, squared is $9$.
For $y$: half of $4$ is $2$, squared is $4$.
Add both to both sides: $(x^2 - 6x + 9) + (y^2 + 4y + 4) = 12 + 9 + 4$
Factor each group: $(x - 3)^2 + (y + 2)^2 = 25$
Center: $(3, -2)$, radius: $5$.
Angles in a circle
Central angle
Vertex at the center. Sides are radii.
Equals its intercepted arc in degrees.
EXAMPLE
A central angle measures $80°$. What is the arc it intercepts?
$80°$ — central angle equals its arc.
Inscribed angle
Vertex on the circle. Sides are chords.
Half the central angle that subtends the same arc.
EXAMPLE
An inscribed angle measures $25°$. What's the corresponding central angle?
Twice the inscribed angle: $50°$
Arc length & sector area
Arc length
$$\frac{\text{arc}}{C} = \frac{\theta}{360°}$$
arc = arc length
C = circumference $= 2\pi r$
θ = central angle in degrees
C = circumference $= 2\pi r$
θ = central angle in degrees
The arc is the same fraction of the circumference as the angle is of $360°$.
EXAMPLE
A circle has radius $6$ and a central angle of $60°$. Find the arc length.
$\dfrac{\text{arc}}{2\pi(6)} = \dfrac{60}{360}$
$\text{arc} = \dfrac{1}{6}(12\pi) = $ $2\pi$
Sector area
$$\frac{\text{sector}}{A} = \frac{\theta}{360°}$$
sector = sector area
A = total area $= \pi r^2$
θ = central angle in degrees
A = total area $= \pi r^2$
θ = central angle in degrees
A sector is a "pizza slice" of the circle. Same proportional setup as arc length.
EXAMPLE
A circle has area $36\pi$ and a $90°$ central angle. Find the sector area.
$\dfrac{\text{sector}}{36\pi} = \dfrac{90}{360}$
$\text{sector} = \dfrac{1}{4}(36\pi) = $ $9\pi$
Tangent lines
Radius is perpendicular to the tangent at the point of contact
Where a tangent line touches the circle, draw the radius to that point. That radius forms a 90° angle with the tangent. Almost always sets up a Pythagorean problem.
EXAMPLE
A tangent line touches a circle at point $P$. The radius from center $O$ to $P$ is $5$. Another point $Q$ on the tangent line is $12$ units from $P$. How far is $Q$ from $O$?
Triangle $OPQ$ has a right angle at $P$, so use Pythagorean:
$OQ^2 = 5^2 + 12^2 = 169 \ \Rightarrow \ OQ = $ $13$