NOTE: This post is about the OLD SAT (pre-2016). The current SAT DOES require you to know things like circle equations and trigonometry. Because the current SAT will also still ask questions like the ones discussed in this post, though, you may still find the information below useful.

Circles can be difficult to deal with, especially if you’re approaching them from a very mathy perspective. Approach them from the perspective of someone who grasps that the SAT is not a math test, and you’re going to have a much easier time.

Here’s what you DON’T need on the SAT:

• Polar coordinates
• Cartesian circle equations (like [xh]2 + [yk]2 = r2)
• Any other kind of circle equations
• Trigonometry

Get those things right out of your head. They’ll only overcomplicate things for you, obfuscating the solution.

On the SAT, you only need to know a few things about circles. Let’s start with what they tell you in the beginning of every section:

• The area of a circle can be calculated using A = πr2.
• For circumference, use C = 2πr (or C = πd, as some prefer–potayto/potahto).
• The number of degrees of arc in a circle is 360.

You also might (rarely) need to know, but aren’t told:

And now here’s what they don’t tell you that you’ve got a very good chance of needing to know:

Many of the most arduous circle problems are actually ratio problems! Think of them this way, and you’ll have a much easier time.

Areas of sectors and lengths of arcs can be calculated using ratios:

We can represent the various parts of this circle with the following ratio:

$\inline&space;\dpi{300}&space;\fn_jvn&space;\frac{part}{whole}=\frac{part}{whole}$

In other words:

$\inline&space;\dpi{300}&space;\fn_jvn&space;\frac{Degree\:Measure\:of\:Central\:Angle}{360^{\circ}}=\frac{Arc\:Length}{Circumference}=\frac{Area_{sector}}{Area_{whole}}$

If you can complete just one of these fractions (for example, if you know the ratio of arc length to circumference), you can figure out everything else. Often all you’ll get are the radius and a central angle, but as you’re about to see, that’s plenty.

##### Example:

Let’s use these ratios to find the area and perimeter of PacMan:

First, calculate the area and circumference of the whole circle (PacMan with his mouth closed):

Awhole = π(3)2

Awhole = 9π

C = 2π(3)

C = 6π

Now we can use the angle of PacMan’s gaping maw to find what we’re looking for. Note that since PacMan’s mouth is open at a 40° angle (which is empty space), we’re actually looking for the part of the circle that is 320°. Let’s tackle his area first:

$\tiny&space;\inline&space;\dpi{300}&space;\fn_jvn&space;\frac{320^\circ}{360^{\circ}}=\frac{A_{PacMan}}{A_{whole}}$

Substitute in our value for Awhole (are you giggling yet?):

$\tiny&space;\inline&space;\dpi{300}&space;\fn_jvn&space;\frac{320^\circ}{360^{\circ}}=\frac{A_{PacMan}}{9\pi}$

Simplify the fraction on the left:

$\tiny&space;\inline&space;\dpi{300}&space;\fn_jvn&space;\frac{8}{9}=\frac{A_{PacMan}}{9\pi}$

So we’ve solved for PacMan’s area:

$\tiny&space;\inline&space;\dpi{300}&space;\fn_jvn&space;8\pi=A_{PacMan}$

The process for his perimeter is very similar. Now, instead of looking for a part of the whole area of the circle, though, we’re looking for a part of its circumference:

$\tiny&space;\inline&space;\dpi{300}&space;\fn_jvn&space;\frac{320^\circ}{360^\circ}=\frac{Arc\:Length}{C}$

Substitute and simplify:

$\tiny&space;\inline&space;\dpi{300}&space;\fn_jvn&space;\frac{8}{9}=\frac{Arc\:Length}{6\pi}$

Solve:

$\tiny&space;\inline&space;\dpi{300}&space;\fn_jvn&space;Arc\:Length=\frac{16}{3}\pi$

But wait! We’re not quite done yet. That’s just the arc length around PacMan. To find his perimeter, we have to add the top and bottom of his mouth, too.

$\tiny&space;\inline&space;\dpi{300}&space;\fn_jvn&space;P_{PacMan}=\frac{16}{3}\pi+6$

There we go. Circle questions: not so bad, right? Don’t panic, just work with the ratios.

##### Try it, dawg.

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Also try this previously posted challenge question (post contains explanation).

shivam patel says:

How do you do number 16?

edit: never mind

The fact that it’s a square in the middle means that each circle is missing 90°. So you’re looking for 270/360 (or 3/4) of each circle.

Michael says:

Hi, can you please explain number 19?
Also for number 16, how would we know whether to find the circumference or the area? For example, I found the area instead of the circumference and it gave me 3π, not the 6π I was looking for.

I think, for #16, you’re only counting one side of the figure. You want the TOTAL area of the shaded regions. Each shaded region alone is 3π. Both of them together is 6π.

As for #19, you know the radius is 6, so you know the total circumference of the circle is 12π. That means the arc you’re given, π, is only 1/12 of the circle. Solve this ratio to get the angle:

π/(12π) = x/360
30 = x

serena says:

how do we know if its 1/12 of the circle ? I’m a little confused

i got to pi/12pi = x/360 but how do you get 30 from that?

Dylan says:

Why isn’t 16 D? I’m having a bit of trouble meshing the shaded regions post with this one.

Because if those circles were closed, they would overlap. So you can’t simply add up the total areas of each circle and subtract the square, or you count the overlapping portion twice. Instead, use part/whole ratios to find the area of 3/4 of the circle…

Dylan says:

So area of part/4pie (area of whole) = 270/360 is the right way to approach it?

Tarsun says:

I still don’t understand how number 16 is 6 pi. I understand you’re supposed to do 270/360 = x/4 pi. I got 3 pi as a result but how do you get 6 pi from that?

Right, but that 3 pi only accounts for one of the two circle sectors. So you have to double it to count all the shaded regions.

Siri says:

For #16, I don’t understand why you can’t simply find the area of the two circles and subtract the area of the square from it??

Reem says:

I understood how to solve both questions, but how would you solve this type of question? In the circle above point O is the center and AB is the diameter. If the length of arc ACB is 4pi, what is the length of arc AED? The radius wasn’t given so I didn’t know how to solve it.

Arc ACB is 180 degrees, and arc AED is 45 degrees. 45 is 1/4 of 180, so arc AED is 1/4 of the length of arc ACB. The answer is π.