Circles on the GMAT function like any other GMAT quant topic: the list of “knowledge bits” you need is short, but the questions creatively combine and/or disguise these few “knowledge bits” to create complex problems.

In this first article, we will discuss the most basic properties of circles and the formulas that relate to these properties. The properties in question are area, circumference, and radius/diameter.

**Circle Properties: Radius & Diameter **

I treat radius (r) and diameter (D) together because they essentially express the same thing, and because the relationship between them is so simple.

Diameter tells you how “wide” a circle is at its widest point. If you draw a straight line all the way across a circle, hitting the center on the way, that line is a diameter, and the length of that line is the diameter of the circle. The radius of the circle is half of this line – or any straight line drawn from the center of the circle to a point on its edge.

This is the most basic property of a circle. The two next most basic properties involve bringing in a specific number, one of the most famous numbers in mathematics: pi.

**Circle Properties: Pi**

Pi is the name of a Greek letter that looks (in its lowercase form) like this: π.

The value represented by this character is irrational (the numbers after the decimal never stop), but for most purposes and for the GMAT, 3.14 is enough. If you happen to be working with fractions or if you prefer fractions to decimals, 22/7 is a rather precise way to express pi. Rounded to the thousandths place, pi is 3.142, and 22/7 is 3.143. That’s close enough for jazz and certainly close enough for GMAT quant.

Interestingly – and for reasons relating to math beyond what is required for the GMAT – this same value can be combined with the radius/diameter to calculate both the circumference (C) and area (A) of the circle. Circumference is the distance around the circle (its perimeter), and area is the space inside the circle.

C = π*D or C = 2*π*r

A = π*r²

The two options for the circumference equation are, in fact, equivalent, since D = 2r. You may be wondering why mathematicians split the 2 and the r around the π, instead of just saying π2r. One reason is simply the conventions that have developed for algebraic notation; it “looks wrong” to have the 2 in the middle after the π. But another reason is a (non-GMAT math) connection between the circumference and area equations. Both equations have a π, an r, and a 2. In the area equation, the 2 functions as an exponent on the r. In the circumference equations, it functions as a coefficient multiplying the expression. Area is pi times the square of the radius. Circumference is pi times the radius doubled.

**Examples:**

1. If a radius is 3, what’s the area of the circle?

A = π*r²= π*3² ≈ 28.27433

2. If the area of a circle is 25π, what is the diameter of the circle?

A = π*r²

r² = 25*π÷π

r = √25 = 5

3. If a radius is 4, what’s the circumference of the circle?

C = 2*r*π

C = 2*4*π

C = 8*π = 25.13274

**Official GMAT Problem**

In all likelihood, none of this looks new. But like we said at the beginning, GMAT quant can get creative with common mathematical knowledge. Take a look at this official GMAT problem, and try to answer it before reading on:

In the figure shown, if the area of the shaded region is 3 times the area of the smaller circular region, then the circumference of the larger circle is how many times the circumference of the smaller circle?

A. 4

B. 3

C. 2

D. √3

E. √2

**Understanding the Problem **

The only properties at play here are area and circumference, perhaps with radius/diameter as a “bridge” between the two, but the answer to the problem may not be immediately obvious. Part of this is due to a common GMAT technique – the removal of numbers to make the problem more abstract.

This problem disguises the relationship between the two circles by referring not to the area of each circle, but to the area of the shaded region. A helpful preliminary deduction is that if, as the problem says, the area of the shaded region is 3 times the area of the smaller circular region, *then the area of the larger circle is 4 times the area of the smaller circle* (comprising the 3 “parts” in the shaded region and the 1 “part” in the smaller circle).

So we have the factor relating the circles’ areas: 4. But we were asked about circumference, not area. How do we get from area to circumference? Well, both the area and circumference equations have the radius in them, so one option is to pick two values for area, with one being 4 times the other, solve for the radius of each circle, and then plug each of these radius values into the circumference equation.

**Solving the Problem **

Let’s say the area of the larger circle is 16, and the area of the smaller circle is 4. Since π is common to all the circle equations, in this case it is irrelevant, and we should just remove it instead of keeping it as “dead weight” to move around algebraically.

Large Circle Small Circle

A = π*r² A = π*r²

16 = π*r² 4 = π*r²

r = 4 r = 2

So when the areas of two circles are related by a factor of 4, their radii (plural for radius) are related by a factor of 2: that is, the square root of 4. This makes sense, since radius is a linear value and area is a square value. This is just like how when you double the length of the side of a square, you quadruple its area (since 2² = 4). When you triple the length of the side of a square, you make its area nine times what it was before (since 3² = 9). The factor of increase for area *is the square of* the factor of increase for the linear measure.

C = 2*π*r

C = 2*π*4

C = 2*π*2

Since we are only concerned about the factor relating the circumferences of the circles, we can ignore whatever is common to both (2*π), leaving us with 4 and 2. Since 4 is twice as much as 2, the circumference of the larger circle is twice the circumference of the smaller circle, and the correct answer choice is **C**.

**The Final Step**

That last step (and any real calculation) is avoided with two insights:

**There is a square relationship between area and radius.**Since the area of the larger circle is 4 times that of the smaller circle, the radius of the larger circle is*the square root of 4*(2) times the radius of the smaller circle.**There is a linear relationship between circumference and radius.**Since the radius of the larger circle is twice that of the smaller circle, the circumference of the larger circle is also twice that of the smaller circle.

When you move beyond rote memorization and *understand* the circle equations, these insights happen naturally and lead you to the correct answer choice quickly, with virtually no calculation.

This concludes the first article in our series on the GMAT’s treatment of circles. Next, we will dive into what happens when you use something called a *central angle* to mark off only a part of the area (a sector) and circumference (an arc) of a circle.

**Contributor: ***Elijah Mize (Apex GMAT Instructor) *