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## Problem solving with distance on the coordinate plane

# Points inside/outside/on a circle

CCSS.Math:

## Video transcript

- [Voiceover] A circle is
centered at the point C, which has the coordinates negative one, comma, negative three. And has a radius of six. Where does the point P, which has the coordinates negative six,
comma, negative six, lie? We have three options. Inside the circle, on the
circle, or outside the circle. And 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. So we say that is its radius. 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 a set of all points that are exactly six units away from the center. So 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. So the key is, is let's
find the distance between these two points. If the distance is less than six, inside, distance equals six, we're on the circle, distance more than six, we
are outside of the circle. So let's do that. So if we wanted to find,
well there's 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 X squared plus
our change in Y squared. So what is our change in X? So our change in X, if we view C as our starting point and P as our end point, and we could do it either way, our change in X, our change in X is negative six minus negative one. So negative six minus negative one. And we're going to square it. So what we have inside
here, that is change in X. So we're taking our change in X squared, and then plus our change in Y squared. So we are going, we're going from negative three to negative six. So our change in Y is negative
six minus negative three. Negative six minus negative three, and we're going to square everything. So that is our change in
Y inside the parentheses, and we're going to square it. So this is equal to, this is equal to negative six, negative
six plus positive one, is one way to think about it, so this is negative five squared. And then this is negative six plus three. So plus negative three squared. And once again, you can see, our change in X is negative five. We go five lower in X, and
we're going three lower in Y. So we're changing Y, is negative three. So this is equal to the square root of 25, square root of 25 plus nine. Square root of 25 plus
nine, which is equal to the square root of 34. Now, the key is, is the square root of 34 less than six, greater
than six, or equal to six? Well, we know that six is equal to the square root of 36,
so the square root of 34 is less than the square root of 36, so 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 six. These are the square root of 36, is six. And so 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.