If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Features of a circle from its expanded equation

Sal finds the center and the radius of a circle whose equation is x^2+y^2+4x-4y-17=0, and then he graphs the circle. Created by Sal Khan.

Want to join the conversation?

  • winston baby style avatar for user Baby Winston
    In the first equation, why does Sal make the y^2 positive?
    (39 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user mirghanimohamed274
    i don't get why you add the 4 in ?
    (16 votes)
    Default Khan Academy avatar avatar for user
    • male robot hal style avatar for user Girish K J
      because, we have to add 4 to make it a square or to make it in the form of (a+b)^2.If you add 4 to (x^2 + 4x):
      =(x^2 + 4x) +4
      =(x^2 + 4x + 4) = (x^2 + 2x + 2x +4)
      =[x(x+2) + 2(x+2)]
      =(x+2) (x+2) = (x+2)^2
      this is how a quadratic equation is solved. we have to bring it into (x-h)^2 form which is a part of (x-h)^2 + (y-k)^2 = y^2
      (25 votes)
  • piceratops seed style avatar for user rpappu360
    Why is it not -y^2? You put positive y^2 In the above video
    (7 votes)
    Default Khan Academy avatar avatar for user
  • duskpin ultimate style avatar for user Victor Chan
    So how would you solve an equation if it goes x^2+y^2= C? Would the center just be (0,0) and radius the square root of C?
    (5 votes)
    Default Khan Academy avatar avatar for user
  • leaf green style avatar for user Evan Campbell
    how can you factor (x^2+2x+2)?
    (2 votes)
    Default Khan Academy avatar avatar for user
    • hopper cool style avatar for user Chuck Towle
      If you want to find the factors of
      You can set the expression equal to 0 making it an equation and then find its roots. Then take (x - first root)(x-second root) and that will be the factors of the original expression.
      While the method works for all equations. you sometimes get irrational factors and/or complex factors using the imaginary number.

      As Sid said, for your expression, you will get a complex factor using the imaginary number.

      Here is how to do it.
      x²+2x+2 = 0 To use "complete the square" method you start my eliminating the constant on the left by subtracting the constant form both sides.
      Now to "complete the square" on the left, you need to
      take the coefficient next to the x divided it by 2 and then square it.
      You have a 2x in the equation, so divide the 2 by 2 which is 1 and square it. 1² is still 1. So add 1 to both sides.
      x²+2x+1 = -2+1
      x²+2x+1 = -1 Now factor the expression on the left
      (x+1)(x+1) = -1 Which is
      (x+1)² = -1 Now take the square root of each side.
      x+1 =±√(-1) The √(-1) is the imaginary number i or more correctly i² = -1 so
      x+1 = ±i Now subtract 1 from both sides
      x=-1±i The ± means you have two answers so you could write it as
      x=-1+i and x=-1-i
      These are your two roots. To find the factors you need to subtract the roots from x
      (x-first root)(x-second root)
      And putting in the roots you have found
      (x-(-1+i))(x-(-1+i)) Distribute the negative signs

      So that is how you find that
      x²+2x+2 factors into
      just as Sid said,

      I hope that is of some help.
      (10 votes)
  • piceratops tree style avatar for user Lakshya
    What did Sal mean when he said the center is the point (a,b) that essentially sets both (x-a)^2+(y-b)^2 equal to zero??
    (3 votes)
    Default Khan Academy avatar avatar for user
  • male robot donald style avatar for user sanket
    Is it possible to have a negative y^2 term? If yes, then how do you solve such kind of equations? I mean won't the negative coefficient make it more confusing? (eg. x^2 + 2x -y^2 =0) How would you solve this problem ( that is if the y^2 term can have a negative coefficient)?
    (4 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user jocrui608
    how would I be able to solve
    (2 votes)
    Default Khan Academy avatar avatar for user
  • leaf green style avatar for user paulhoughton.uk
    I'm doing Geometry but didnt do Algebra yet (mostly because that seems to be the order it comes in KA)
    The subject completing the square suddenly appears here in Geometry but the videos are in the algebra section I didnt do yet
    Am I missing a simple intro to completing the square enough to do this problem without working through algebra first?
    Thanks for any guidance
    (5 votes)
    Default Khan Academy avatar avatar for user
    • starky sapling style avatar for user ralph117
      I would advice learning how to solve the square first, because that's used over and over again in Conic Sections. Conic Sections are technically speaking Pre-Calc, not geometry. That's why we use solving the square. Last year, my math class was an accommodation of both geometry and algebra to make math easier (we did circles back then). And believe me, it was a lot easier that way. I would just go ahead and watch those videos, the concept really isn't that hard to grasp and if you do understand it, then you'll be ahead of your classmates!
      (1 vote)
  • blobby green style avatar for user amyyun1218
    I have question for you why is it you use the long equation of this next lesson about the equation of circle?
    (3 votes)
    Default Khan Academy avatar avatar for user

Video transcript

We're asked to graph the circle. And they give us this somewhat crazy looking equation. And then we could graph it right over here. And to graph a circle, you have to know where its center is, and you have to know what its radius is. So let me see if I can change that. And you have to know what its radius is. So what we need to do is put this in some form where we can pick out its center and its radius. Let me get my little scratch pad out and see if we can do that. So this is that same equation. And what I essentially want to do is I want to complete the square in terms of x, and complete the square in terms of y, to put it into a form that we can recognize. So first let's take all of the x terms. So you have x squared and 4x on the left-hand side. So I could rewrite this as x squared plus 4x. And I'm going to put some parentheses around here, because I'm going to complete the square. And then I have my y terms. I'll circle those in-- well, the red looks too much like the purple. I'll circle those in blue. y squared and negative 4y. So we have plus y squared minus 4y. And then we have a minus 17. And I'll just do that in a neutral color. So minus 17 is equal to 0. Now, what I want to do is make each of these purple expressions perfect squares. So how could I do that here? Well, this would be a perfect square if I took half of this 4 and I squared it. So if I made this plus 4, then this entire expression would be x plus 2 squared. And you can verify that if you like. If you need to review on completing the square, there's plenty of videos on Khan Academy on that. All we did is we took half of this coefficient and then squared it to get 4. Half of 4 is 2, square it to get 4. And that comes straight out of the idea if you take x plus 2 and square it, it's going to be x squared plus twice the product of 2 and x, plus 2 squared. Now, we can't just willy-nilly add a 4 here. We had an equality before, and just adding a 4, it wouldn't be equal anymore. So if we want to maintain the equality, we have to add 4 on the right-hand side as well. Now, let's do the same thing for the y's. Half of this coefficient right over here is a negative 2. If we square negative 2, it becomes a positive 4. We can't just do that on the left-hand side. We have to do that on the right-hand side as well. Now, what we have in blue becomes y minus 2 squared. And of course, we have the minus 17. But why don't we add 17 to both sides as well to get rid of this minus 17 here? So let's add 17 on the left and add 17 on the right. So on the left, we're just left with these two expressions. And on the right, we have 4 plus 4 plus 17. Well, that's 8 plus 17, which is equal to 25. Now, this is a form that we recognize. If you have the form x minus a squared plus y minus b squared is equal to r squared, we know that the center is at the point a, b, essentially, the point that makes both of these equal to 0. And that the radius is going to be r. So if we look over here, what is our a? We have to be careful here. Our a isn't 2. Our a is negative 2. x minus negative 2 is equal to 2. So the x-coordinate of our center is going to be negative 2, and the y-coordinate of our center is going to be 2. Remember, we care about the x value that makes this 0, and the y value that makes this 0. So the center is negative 2, 2. And this is the radius squared. So the radius is equal to 5. So let's go back to the exercise and actually plot this. So it's negative 2, 2. So our center is negative 2, 2. So that's right over there. X is negative 2, y is positive 2. And the radius is 5. So let's see, this would be 1, 2, 3, 4, 5. So you have to go a little bit wider than this. My pen is having trouble. There you go. 1, 2, 3, 4, 5. Let's check our answer. We got it right.