 So now that we know all there is to know about graphing straight lines, or do we, we can try to consider graphing other objects. Let's take a look at circles. In order to translate the geometric object of a circle into some algebraic equation, we have to go back to basics. So what's the concept of a circle? Well, we have some point and some radius, and the circle is defined by all points that are that distance away from the center. We'll get an equation by translating these ideas into the language of algebra. So given a center point, which will identify by its coordinates h, k, and a radius, r, a circle is the set of all points that are r units away from the center h, k. Now, we'd like an equation that tells us something about the coordinates of any point on the circle. So suppose my point x, y is on the circle. Then it must be the case that the distance between this point x, y, and the center h, k has to be r. But we can use our distance formula to calculate what that distance is. Actually, we'll take one more step for a variety of reasons, most of which you'll understand when you get to calculus, it's inconvenient to have a square root as part of an equation. So if we square both sides, we can get rid of that square root and have our equation in a somewhat nicer form. And since h, k, and r are specific values that identify the center and radius of a circle, this is an equation in two variables that can be the equation for the circle. And putting everything together, this tells us the following theorem. The circle with center h, k, and radius r has the equation x minus h squared plus y minus k squared equals r squared. Now, if the universe is a kind and gentle place, you'll always be given the center and radius of a circle. And just like being given the point and the slope of a line, once you have that information, you can immediately write down the equation of the circle. We need the coordinates of the center and the radius. And of course, you can imagine how often you're given the center and radius of a circle and told to write the equation. That's right, except for this one problem, likely never. So one variation is to find the center and radius of the circle with the equation this, and that looks nothing like the equation of a circle. And since this isn't in this form, we should put it in this form using some algebra. So notice that our equation for the circle has the form an expression of x squared plus an expression of y squared equals the square of some real number. So what that suggests is that we can put our equation into this form by completing the squares on x and y separately. So what we'll do is we'll separate our x terms and our y terms and give us a little bit of space to work in. Now we want x squared minus 6x to be the start of a perfect square. So for the square beginning with x squared minus 6x, we note that x is equal to x squared. Minus 6x is 2 times x times negative 3. And so we can complete the square by adding negative 3 squared or 9. And because I have an equation, if I add 9 to one side, I have to add 9 to the other side. How about my y terms? So for the square beginning with y squared plus 8y, we note that y is equal to y squared. 8y is 2 times y times 4. So we can complete the square by adding 4 squared or 16. So we'll add 16 to one side and 16 to the other side. So now my x terms, x squared minus 6x plus 9, those form a perfect square, x minus 3 squared. My y terms, y squared plus 8y plus 16, those also form a perfect square, namely. And my right hand side, well, I could do the arithmetic. That's 25 plus 9 plus 16. And now my equation is in a familiar looking form. And once it's in this form, I can read off the center hk and the radius. We can make this a little bit more obvious. We want our x terms to be x minus something squared. Well, it is. We'd like our y terms to be y minus something squared. Well, since it's y plus 4, I'll rewrite it as y minus negative 4. And finally, we want the thing over on the right hand side to be the square of something. So 50 can be rewritten as, and so I can compare our equation to the equation of a circle, and we see that our center is going to be located at 3 negative 4, and the radius will be square root of 50. At this point, you know everything you need to know to write the equation of any circle. And here a quote from Descartes is appropriate. When he first described this relationship between curves and algebraic expressions, he said that I shall not stop to explain this in more detail because I should deprive you of the advantage of training your mind by working over it. So if you want to do yourself a favor, skip this video, and simply work out how to find the equation of any given circle given any bits of information. For those of you who are still watching, here's a few more examples. So let's try to find the equation of the circle with center 5, 3 that passes through the point 1 negative 4. So remember, we can write the equation of a circle if we have the center and radius. Well, we have the center, but we don't have the radius. So how can we get the radius? Well, the radius of the circle will be the distance between the center of the circle and any point on the circle. Well, we know the center, we know a point on the circle, and we know how to find the distance between them. So the radius will be square root of 65, so an equation of the circle will be... What if we don't know the center of the circle? So we may want to try to find the equation of the circle whose diameter has endpoints negative 3, 5, and 1 negative 1. If only we had the center and radius of the circle. But wait, the center is the midpoint of the diameter, so we can find its location by finding the midpoint of the diameter. The midpoint of the diameter will be negative 1, 2. The radius is the distance between the center and any point on the circle. We know where the center is, we know two points on the circle, so we can find the distance between the center, negative 1, 2, and either point. We'll use the point 1, negative 1. This tells us the radius is square root of 13, so we have the center, we have the radius, and now we can write the equation of the circle.