If the radius of a circle that centers at the origin is 5, how many po

There is a necessity to understand some of the verbiage in this question in order to be able to answer it properly. Firstly, a circle that is centered at the origin is centered at point {0,0}. The radius is 5, which means we know the diameter (2*r), the circumference (2*π*r) and the area (π * r^2). However, none of that information really helps us to answer this question. We are interested in how many points on the circle have integer coordinates. Quite simply, a circle has an infinite number of discrete points, so it’s easier to answer this question in the reverse: For each integer coordinate, is that point on the circle?

Let’s start with the obvious ones. The point {5,0} has to necessarily be on the circle. If the origin is {0,0} and the radius is 5, then not only must point {5,0} be on the circle, but so must point {-5,0}. The circle extends in all four directions, so we cannot forget the negative values. Similarly, the points {0,5} and {0,-5} will also be on the points, effectively covering the four cardinal points from the original circle. The circle could look something like this:


After solving for these four points, we must evaluate whether other integer coordinates could be on the circle. One thing should be clear: if the radius is 5, then any integer point above 5 will necessarily not be on the circle, as it is beyond the reach of our radius. We’ve already covered zero, so the only options we have left are one, two, three and four. Of course all of these numbers have negatives and can be considered on either the x or y axis, but still we have a finite number of possibilities to consider.

Another important thing that might not be as obvious is that the answer to this question will necessarily be a multiple of four. Given that a circle extends in all directions by the same distance, it is impossible for point {x, y} to be on the circle and for points {x,-y}, {-x,y} and {-x,-y} to not also be on the circle. This is an important property of all circles and one of the reasons they’re so common in everything from architecture to cooking (and to alien crop circles, if you believe in that). This rule also guarantees that any answer choice that’s not a multiple of four can be eliminated. We can thus eliminate answer choice D (15).

How do we go about finding other points on the circle? (Why am I asking rhetorical questions?) By using the Pythagorean Theorem, of course! Any point on the circle naturally forms a right angle triangle with the radius as the hypotenuse, and the radius is always five. Therefore, if the two other sides can be formed out of integers, we have a point on the circle with integer coordinates. The graph below will highlight this principle:


Since the Pythagorean Theorem states that the squares of the sides will be equal to the square of the hypotenuse, we only need to look for numbers that satisfy the equation a^2 +b^2 = r^2. And given that r is 5, r^2 must always be 25. So if we plug in a as one, we find that 1 + b^2 = 25. This gives us b^2 = 24, or b = √24, which is not an integer. We only have to plug this in three more times, so there’s no reason not to try all the possibilities. If a = 2, then we get 4 + b^2 = 25. The value of b would be √21, which again is not an integer value.

If a = 3, however, we quickly recognize the vaunted 3-4-5 triangle, as 9 + b^2 = 25, meaning b^2 is 16 and therefore b is 4. This means that the points {3,4}, {-3,4}, {3,-4} and {-3,-4} are all on the circle. We’ve brought the total up to 8, but we’re not done. The final value is when a equals four, which will again work and bring in the converse of the last iteration: {4,3}, {-4,3}, {4,-3} and {-4,-3}. These values are distinct from the previous ones, so we now have a total of 12 points. We’ve already checked five, so we can stop here. The answer to this question is answer choice C. There will be 12 distinct values with integer coordinates, as crudely demonstrated below (or on any analog clock).


In geometry, even if it feels like you have to constantly commit more rules to memory, remember that the rules are not nearly as important as knowing how to apply them. This problem can be solved with just the Pythagorean Theorem and a little elbow grease (or a graphing calculator). The GMAT is very much a test of how you think, not of what you know. If you think about geometry problems as cases that must be solved, or obstacles to be overcome, you’ll be in good shape to solve them.