How do I remember whether to add or subtract when reading the center?

The standard form uses subtraction: (x-h) and (y-k). So whatever number appears, flip its sign to get the center coordinate. (x+3) means h=-3; (x-3) means h=3. The center is always the value that makes each squared term equal zero.

When does the SAT use radians instead of degrees for arcs?

Either can appear. Check the angle: if it contains or is described as radians, use s=r and 1/2r^2. If the angle is a plain number of degrees, use the fraction-of-360 formulas. Never mix them — convert first using 180^= radians.

Do I always have to complete the square, or can I sometimes skip it?

If the equation is already in standard form (x-h)^2+(y-k)^2=r^2, just read off the values. You only complete the square when given the general form x^2+y^2+Dx+Ey+F=0, where the center and radius are hidden.

Should I leave my answer with or convert to a decimal?

Match the answer choices. On multiple-choice questions the SAT usually keeps symbolic, so 49/5 is the intended form. For student-produced responses, enter a decimal only if the answer isn't a clean fraction, and round according to the problem's instructions.

DSAT-4.4

Circles: Equations, Arcs & Sectors

Master Digital SAT circles: complete the square to find center and radius, and compute arc length and sector area from central angles.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on Circles: Equations, Arcs & Sectors, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start Circles: Equations, Arcs & Sectors free

What this lesson covers

Circle questions on the Digital SAT come in two flavors: algebraic problems where you convert a messy equation into standard form to read off the center and radius, and geometric problems where a central angle unlocks arc length or sector area. Both reward you for knowing a small set of formulas cold and applying them without hesitation.

In this lesson you'll learn to complete the square from the general form of a circle, interpret the standard form

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

, and use proportional reasoning to slice off arcs and sectors. These skills show up in the calculator and no-calculator sections alike, so speed and accuracy both matter.

Standard Form and What It Tells You

The standard equation of a circle is

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

Here

(h,k)

is the center and

r

is the radius. The single most common mistake is sign confusion: because the form uses subtraction, the equation

(x-3)^2+(y+5)^2=16

has center

(3,-5)

, not

(3,5)

, and radius

\sqrt{16}=4

, not 16.

Read each piece carefully:

Equation piece	Interpretation
$(x-h)^2$	$x$ -coordinate of center is $h$
$(y-k)^2$	$y$ -coordinate of center is $k$
right side $=r^2$	radius is $\sqrt{r^2}$

The SAT loves to give you the standard form directly and ask for the radius, the center, or a point on the circle. If a question gives the center and radius and asks for the equation, just plug in — remember to square the radius on the right side. A circle with center

(-2,4)

and radius 6 has equation

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

Completing the Square from General Form

The general form of a circle is

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

To find the center and radius you must convert this into standard form by completing the square separately for the

x

terms and the

y

terms.

The procedure: group the

x

terms together and the

y

terms together, move the constant to the right side, then for each variable take half of its linear coefficient, square it, and add that value to both sides.

For example, with

x^2+6x

, half of 6 is 3, and

3^2=9

, so

x^2+6x+9=(x+3)^2

. You added 9, so you must add 9 to the other side too.

Step	Action
1	Group $x$ 's and $y$ 's; move constant right
2	Half the $x$ coefficient, square it, add both sides
3	Half the $y$ coefficient, square it, add both sides
4	Factor each group into $(x-h)^2$ and $(y-k)^2$
5	Read center $(h,k)$ and radius $\sqrt{r^2}$

Watch out when the coefficients are odd: half of 5 is

2.5

, and

2.5^2=6.25

. The arithmetic is still fine — just keep the fractions.

Arc Length from a Central Angle

An arc is a portion of the circle's circumference. A central angle is an angle with its vertex at the center; it cuts off an arc. The key idea is proportion: the arc is the same fraction of the whole circumference as the central angle is of a full rotation.

In degrees, if the central angle is

\theta

degrees:

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

The full circumference

2\pi r

is scaled down by the fraction

\frac{\theta}{360}

. A

90^\circ

angle gives one quarter of the circumference; a

180^\circ

angle gives half.

The Digital SAT sometimes uses radians. In radians, arc length has an even cleaner form:

s=r\theta

where

\theta

is in radians. This works because a full circle is

2\pi

radians, so

\frac{\theta}{2\pi}\cdot 2\pi r=r\theta

. Know both versions and check which unit the problem uses. A common trap is plugging a degree measure into

s=r\theta

— always confirm the angle is in radians first.

Sector Area and the Radian Connection

A sector is a pie-slice region bounded by two radii and an arc. Just like arc length, sector area is the same fraction of the whole circle's area as the angle is of a full turn.

In degrees:

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

In radians:

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

Both give the same answer when you convert consistently. Notice arc length and sector area share the same fraction

\frac{\theta}{360}

— only the base quantity changes (circumference for arcs, area for sectors).

Quantity	Degree formula	Radian formula
Arc length	$\frac{\theta}{360}\cdot 2\pi r$	$r\theta$
Sector area	$\frac{\theta}{360}\cdot \pi r^2$	$\frac{1}{2}r^2\theta$

To convert between units, use

180^\circ=\pi

radians, so multiply degrees by

\frac{\pi}{180}

to get radians. The Digital SAT frequently asks you to express an angle in radians given an arc length, or to work backward from a known sector area to find the radius or angle. Set up the proportion, then solve algebraically.

How the Exam Combines These Ideas

The Digital SAT rarely asks a circle question in total isolation. Expect blends: a general-form equation where you complete the square to find the radius, then use that radius in an arc or sector formula. Or a coordinate-geometry setup where you must confirm a point lies on the circle by checking it satisfies the equation.

A frequent question type gives you the equation

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

and asks only for the radius or the coordinates of the center — you complete the square and stop. Another gives the radius and a central angle and asks for arc length or sector area; here you must pick the right formula and match units.

Be alert to the difference between diameter and radius: if a question hands you the diameter, halve it before using any formula. Also remember that the SAT answer choices often leave

\pi

in symbolic form, so don't rush to a decimal unless the problem demands one. Finally, when a figure is drawn, it may not be to scale — trust the given numbers, not your eyes.

Key terms

Standard form of a circle.: The equation $(x-h)^2+(y-k)^2=r^2$ , where $(h,k)$ is the center and $r$ is the radius.
General form of a circle.: The equation $x^2+y^2+Dx+Ey+F=0$ , which must be converted by completing the square to reveal center and radius.
Completing the square.: An algebraic technique that turns $x^2+bx$ into a perfect square $(x+\frac{b}{2})^2$ by adding $(\frac{b}{2})^2$ .
Central angle.: An angle whose vertex is at the center of the circle; it determines the fraction of the circle an arc or sector spans.
Arc length.: The distance along the circle's edge subtended by a central angle, equal to $\frac{\theta}{360}\cdot 2\pi r$ in degrees or $r\theta$ in radians.
Sector.: A pie-slice region of a circle bounded by two radii and an arc; its area is $\frac{\theta}{360}\cdot \pi r^2$ or $\frac{1}{2}r^2\theta$ .
Radian.: An angle measure where a full circle is $2\pi$ radians; convert from degrees by multiplying by $\frac{\pi}{180}$ .

Worked example

The circle in the

xy

-plane is defined by

x^2+y^2-10x+4y-20=0

. What are the coordinates of its center and the length of its radius? Then find the area of the sector formed by a

72^\circ

central angle.

Start by grouping and moving the constant right:

x^2-10x+y^2+4y=20

.

Complete the square for

x

: half of

-10

-5

, and

(-5)^2=25

. Add 25 to both sides.

Complete the square for

y

: half of

4

2

, and

2^2=4

. Add 4 to both sides.

Now the equation is

x^2-10x+25+y^2+4y+4=20+25+4

, which factors to

(x-5)^2+(y+2)^2=49

.

So the center is

(5,-2)

and the radius is

\sqrt{49}=7

.

For the sector, use

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

. With

\theta=72

and

r=7

\frac{72}{360}\cdot \pi (7)^2=\frac{1}{5}\cdot 49\pi=\frac{49\pi}{5}

.

The center is

(5,-2)

, the radius is

7

, and the sector area is

\frac{49\pi}{5}

square units, or about

30.8

Practice questions

The equation of a circle is

(x+3)^2+(y-6)^2=25

. Which of the following is the center and radius of the circle?

Center $(-3,6)$ , radius $5$
Center $(3,-6)$ , radius $5$
Center $(-3,6)$ , radius $25$
Center $(3,-6)$ , radius $25$

Answer: Center $(-3,6)$ , radius $5$

Standard form is

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

. Because the form subtracts

h

and

k

, the term

(x+3)

means

h=-3

and

(y-6)

means

k=6

, giving center

(-3,6)

. The right side is

r^2=25

, so

r=\sqrt{25}=5

, not 25. The trap answers flip the signs of the center or mistake

r^2

for

r

A circle has radius

9

. A central angle measures

\frac{2\pi}{3}

radians. Find the exact arc length it subtends.

Answer: $6\pi$

Since the angle is in radians, use

s=r\theta

directly. Substitute

r=9

and

\theta=\frac{2\pi}{3}

s=9\cdot \frac{2\pi}{3}=\frac{18\pi}{3}=6\pi

. If you had used the degree formula you would first convert

\frac{2\pi}{3}

radians to

120^\circ

, then

\frac{120}{360}\cdot 2\pi(9)=\frac{1}{3}\cdot 18\pi=6\pi

— the same result.

The circle

x^2+y^2+8x-6y+9=0

is graphed in the

xy

-plane. What is the radius of the circle?

Answer: $4$

Group and move the constant:

x^2+8x+y^2-6y=-9

. Complete the square for

x

: half of 8 is 4,

4^2=16

. For

y

: half of

-6

-3

(-3)^2=9

. Add both:

x^2+8x+16+y^2-6y+9=-9+16+9

, so

(x+4)^2+(y-3)^2=16

. The radius is

\sqrt{16}=4

FAQ

How do I remember whether to add or subtract when reading the center?: The standard form uses subtraction: $(x-h)$ and $(y-k)$ . So whatever number appears, flip its sign to get the center coordinate. $(x+3)$ means $h=-3$ ; $(x-3)$ means $h=3$ . The center is always the value that makes each squared term equal zero.
When does the SAT use radians instead of degrees for arcs?: Either can appear. Check the angle: if it contains $\pi$ or is described as radians, use $s=r\theta$ and $\frac{1}{2}r^2\theta$ . If the angle is a plain number of degrees, use the fraction-of-360 formulas. Never mix them — convert first using $180^\circ=\pi$ radians.
Do I always have to complete the square, or can I sometimes skip it?: If the equation is already in standard form $(x-h)^2+(y-k)^2=r^2$ , just read off the values. You only complete the square when given the general form $x^2+y^2+Dx+Ey+F=0$ , where the center and radius are hidden.
Should I leave my answer with $\pi$ or convert to a decimal?: Match the answer choices. On multiple-choice questions the SAT usually keeps $\pi$ symbolic, so $\frac{49\pi}{5}$ is the intended form. For student-produced responses, enter a decimal only if the answer isn't a clean fraction, and round according to the problem's instructions.

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