A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • one year ago

What is the general form of the equation of a circle with its center at (-2, 1) and passing through (-4, 1)?

  • This Question is Open
  1. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Are you trying to find the circumference or radius?

  2. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    (x-h)^2 + (y-k)^2 = r^2 is the standard equation for a circle for starters. (h,k) is the center of the circle; the point given was (-2,1) -2 is point h, and 1 is point k. R^2 is the radius squared. H corresponds with X and K corresponds with K for a side note. So using the points (-2,1) and (-4,1) to find it's radius, the differences between -2 and -4 is 2 points. So 2 is the radius of the circle. So now we got the points (-2,1) and the radius of the circle. Now all we have to do is put it in equation form. (-2, 1) r=2 would become (x-h)^2 + (y-k)^2 = r^2. In the end, it is (x+2)^2 + (y-1)^2=2^2

  3. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.