Points inside/outside/on a circle | Mathematics I | High School Math | Khan Academy

A circle is centered at the point C which has the coordinates -1, -3 and has a radius of six. Where does the point P, which has the coordinates -6, -6, lie? We have three options: inside the circle, on the circle, or outside the circle.

The key realization here is just what a circle is all about. If we have the point C, which is the center of a circle, a circle of radius six. So, let me draw that radius. Let's say that is its radius; it is six units. The circle will look something like this. Remember, the circle is a set of all points that are exactly six units away from that center.

So, that's the definition of a circle; it's the set of all points that are exactly six units away from the center. If, for example, P is less than six units away, it's going to be inside the circle. If it's exactly six units away, it's going to be on the circle, and if it's more than six units away, it's going to be outside of the circle.

The key is, let's find the distance between these two points. If the distance is less than six, we are inside; if the distance equals six, we're on the circle; and if the distance is more than six, we are outside of the circle. So, let's do that.

If we wanted to find—and there are different notations for the distance—well, I'll just write D, or I could write the distance between C and P is going to be equal to—and the distance formula comes straight out of the Pythagorean theorem—but it's going to be the square root of our change in x² plus our change in y².

So what is our change in X? If we view C as our starting point and P as our endpoint (but we could do it either way), our change in X is -6 minus -1, so -6 minus -1 and we're going to square it. So what we have inside here—that is the change in X.

So, we're taking our change in X and then plus our change in y². So, we are going from -3 to -6, so our change in y is -6 minus -3, which is -6 - -3, and we're going to square everything. So, that is our change in y inside the parenthesis, and we're going to square it.

This is equal to -6 plus 1 is one way to think about it, so this is -5² and then this is -6 + 3, so that is -3². Once again, you can see our change in x is 5; we go five lower in X, and we're going three lower in y, so our change in y is 3.

So this is equal to the square root of 25 plus 9. The square root of 25 + 9, which is equal to the square root of 34. Now the key is, is the square root of 34 less than 6, greater than 6, or equal to 6?

Well, we know that 6 is equal to the square root of 36. So, the square root of 34 is less than the square root of 36. I could write the square root of 34 is less than the square root of 36, and so the square root of 34 is less than 6.

Since the distance between C and P is less than six, we are going to be on the inside of the circle. If I somehow got square root of 36 here, then we’d be on the circle, and if I somehow got square root of 37 here or something larger, we would have been outside the circle.