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).