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

Circle equations — Harder example

​Watch Sal work through a harder Circle equations problem.

Want to join the conversation?

  • starky ultimate style avatar for user Caleb Hu
    Why h and k? Standard way of doing it?
    (16 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user Wendy Vickky
    I saw a lot of disheartened and anxious comments while I was preparing for my test so I'm here to say guys keep practicing! you can do it! Khan Academy is such a great resource. I literally only prepared for the SAT for 3 weeks ( 2 weeks for math and one for English) and did not have much hope but I got a great result. you can too! best of luck!
    (23 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user Tiffani.Johnson
    How do u find the radius if it not simple like this problem?
    (10 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user aarohan567
    Best of luck for today everyone ! Do your best !!
    (14 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user basnetrashila
    what is the circumference of a circle whose area is 100 pie
    (6 votes)
    Default Khan Academy avatar avatar for user
    • mr pants orange style avatar for user wruwani612
      20 pie. The equation for the area of the circle is A= (pie)(radius^2). Using that, you can calculate that the radius of the circle is 10. The equation for the circumference of a circle is C=2(pie)(radius) or C=(pie)(diameter). Plug the radius of 10 into an equation and you should find the circumference of the circle to be 20 pie.
      (7 votes)
  • blobby green style avatar for user katieleonard032
    HERE IS THE PROVIDED QUESTION:
    "A circle in the xy-plane contains the points (−1,1), (1,1), and (−1,−1). Which of the following is an equation of the circle?"

    HERE IS THE PROVIDED ANSWER:
    The formula for a circle with a center (h,k) is:
    (x−h)^2+(y−k)^2=r^2


    We could check which formula is satisfied by the three given points.
    Alternatively, let's substitute the three points for (x,y)

    First, substituting (1,1), we have:
    (1−h)^2+(1−k)^2=r^2

    Substituting (−1,1), we have:
    (−1−h)^2+(1−k)^2=r^2

    Finally, substituting (−1,−1), we have:
    (−1−h)^2+(−1−k)^2=r^2


    Next, let's subtract the second equation from the first. We get:
    (1−h)^2−(−1−h)^2=0
    OR
    h^2−2h+1−(h^2+2h+1)=0
    −4h=0
    h=0

    Let's subtract the third equation from the second. We get:
    (1−k)^2−(−1−k)^2=0

    This is the same equation as that above with
    k instead of h. So k=0, and the center of the circle is (0,0). Now, let's find r.

    Since (1,1) is a point on the circle, and the center is (0,0), the radius is squarerootof2.

    The equation of the circle is
    x^2+y^2=2x OR x^2+y^2−2=0



    HERE ARE MY QUESTIONS:

    I don't understand how the answer was found, and I also have a few general questions:
    1. Is it assumed that none three given points are the center of the circle?
    Because I do not know how to write a circle equation without knowing the center. If the center is known to not be one of the three listed points, then it is obviously (0,0), but if the center IS one of the listed points, then it is (-1,1). However, I do not know if the center is included in the listed points or not, so I couldn't figure much else out without that information....

    2. The provided answer talks about subtracting equations from each other like it's a normal thing. I have never encountered a situation where an equation is subtracted form another when dealing with circles... elaborate?
    (3 votes)
    Default Khan Academy avatar avatar for user
    • duskpin ultimate style avatar for user flotar
      Yes, if a point lies ON the circle, then it cannot be its center. That point must be on the circle's edge. I would suggest imagining the coordinates in your head, since the provided points on the circle were very simple. It wouldn't be too difficult to visualize that the center must be (0,0) from the given points.
      (4 votes)
  • blobby purple style avatar for user Veda
    So...based on this video, can it be assumed that when the SAT says "a point on a circle," it is the radius? Like in this case it was 20?
    (3 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user Aaron Cheng
    how do we know the (7, 34) is at the edge of the circle ?
    (2 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user sharvari.rubade2007
    where can we find extra questions for the same topic?
    (3 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user elias
    Elias Torda
    (3 votes)
    Default Khan Academy avatar avatar for user

Video transcript

- [Instructor] A circle in the xy-plane has its center at seven comma 14. If the point seven comma 34 lies on the circle, which of the following is an equation of the circle? Let's just remind ourselves the general form of the equation of a circle. If any of this looks like, looks unusual to you, you haven't seen it before, I encourage you to watch the equation of a circle videos on Kahn Academy. The general form for the equation of a circle, if we have a circle that's centered at the point h comma k, the x-coordinate of the center is h. The y-coordinate of the center is k. If it has a radius of r, so the circle is gonna look something like that. I'm trying to draw it as circular as I can. Man, that's not so circular. I think you get the point. The radius is r. This equation, the equation of this circle, of all the points that are exactly r away from the point h comma k, is going to be x minus the x-coordinate of the center squared plus y minus the y-coordinate of the center squared is equal to the radius squared. Now in this case, we know what our h and k is. H is seven, and k is 14, and we just need to figure out the radius. We figure that out, we will be able to figure out the equation of the entire circle. Now, they tell us that the center's at seven comma 14, so that point right over there is seven comma 14, and then the point seven comma 34 also lies on the circle, so it has the same x-coordinate. Its y-coordinate is just higher, so this might be, that right over there might be the point seven comma 34. It would be right above it. We just increased our y without changing our x. So, seven comma 14 is the center. This point right over here is seven comma 34, lies on the circle, so the circle is gonna look something like this, and so the radius of this circle, what we have to figure out, is just the distance between these two points. These two points, you don't even have to really apply the distance formula or anything like this. The distance here is just our change in y. Our x doesn't change. It's gonna be 34 minus 14. So, the radius here is equal to 20. So, now we know. We know h is equal to seven. K is equal to 14, and r is equal to 20. So, it's gonna be x minus h, x minus seven squared plus y minus k, y minus 14 squared is equal to r squared. 20 squared is 400. So let's see, that is this choice right over here. X minus seven squared plus y minus 14 squared is equal to 400, and we're done.