Circles: Arcs, Sectors, and Equations

The Radius Is the Master Key

Almost every SAT circle question is really a question about the radius. Circumference, area, arc length, and sector area are all built from $r$, and the equation of a circle simply records the center together with $r$. If you can always get back to the radius, the rest is arithmetic. Circle problems split into two families: measurement (circumference, area, arcs, sectors, angles) and coordinate (the equation of a circle). We will handle both, and keep the radius at the center of everything.

Circumference and Area

For a circle of radius $r$, the distance around it — the circumference — is

$C = 2\pi r,$

and the space enclosed — the area — is

$A = \pi r^2.$

Because the diameter is $d = 2r$, you can also write $C = \pi d$. The SAT loves to give you one quantity and ask for another, which almost always means routing through $r$. If a problem tells you the area is $49\pi$, solve $\pi r^2 = 49\pi$ to get $r = 7$, then feed that radius into $C = 2\pi r = 14\pi$. Notice that you rarely need a decimal value of $\pi$: the SAT usually wants the answer "in terms of $\pi$."

Arcs and Sectors Are Just Fractions

An arc is a slice of the circumference, and a sector is a "pizza slice" bounded by two radii and an arc. Both are the same fraction of the whole circle, and that fraction is the central angle over a full turn:

$\text{Arc length} = \frac{\theta}{360^\circ}\cdot 2\pi r, \qquad \text{Sector area} = \frac{\theta}{360^\circ}\cdot \pi r^2.$

A $90^\circ$ sector is exactly one quarter of the circle; a $120^\circ$ sector is one third. The digital SAT also uses radians, where a full turn is $2\pi$ radians. When the angle is in radians the arc length collapses to the tidy formula

$s = r\theta.$

Because $180^\circ = \pi$ radians, you convert by multiplying by $\frac{\pi}{180^\circ}$ (degrees to radians) or $\frac{180^\circ}{\pi}$ (radians to degrees). Keep the units straight: $s = r\theta$ only works when $\theta$ is in radians.

Central Angles versus Inscribed Angles

Two different angles can "look at" the same arc, and confusing them is the single most common circle mistake on the test.

• A central angle has its vertex at the center. It equals the measure of the arc it cuts off.
• An inscribed angle has its vertex on the circle. It equals half the arc it subtends.

So an inscribed angle and a central angle that see the same arc are in a $1:2$ ratio — the inscribed one is half the central one. A famous consequence: any angle inscribed in a semicircle (an inscribed angle that opens onto a diameter) is a $90^\circ$ right angle, because the diameter is a $180^\circ$ arc and half of $180^\circ$ is $90^\circ$.

The Equation of a Circle

A circle centered at $(h, k)$ with radius $r$ satisfies

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

Read the center by flipping the signs inside the parentheses, and take the square root of the right-hand side to get the radius (the right side is $r^2$, not $r$). So $(x-3)^2 + (y+4)^2 = 36$ has center $(3, -4)$ and radius $6$.

When the equation arrives in expanded form — like $x^2 + y^2 - 6x + 8y - 11 = 0$ — you recover the center and radius by completing the square:

(x^2 - 6x) + (y^2 + 8y) &= 11 \\ (x^2 - 6x + 9) + (y^2 + 8y + 16) &= 11 + 9 + 16 \\ (x - 3)^2 + (y + 4)^2 &= 36 \end{aligned}$$ To complete the square in $x$, take half of the $x$-coefficient ($-6 \to -3$), square it ($9$), and add it to **both** sides. Do the same for $y$ (half of $8$ is $4$, squared is $16$). The center is $(3, -4)$ and the radius is $\sqrt{36} = 6$. ## Building a Circle From Its Diameter A slightly harder coordinate problem gives you the two endpoints of a diameter and asks for the equation. The center is the **midpoint** of the diameter, and the radius is **half the distance** between the endpoints. This weaves together the midpoint formula, the distance formula, and the circle equation — a classic SAT synthesis question. ## Common Mistakes to Avoid - Using the *full* arc for an inscribed angle instead of half of it. - Reading the center of $(x-3)^2 + (y+4)^2 = 36$ as $(-3, 4)$ instead of $(3, -4)$. - Reporting $r^2$ as the radius — remember to take the square root. - Adding a completing-the-square constant to the left side but forgetting to add it to the right.