Results 1 to 4 of 4

Math Help - Finding Circle

  1. #1
    Member
    Joined
    Jan 2010
    Posts
    142

    Finding Circle

    I'm having hard time finding equation of a circle with given two points and a radius.

    given points (-2,3) and (2,-1) and radius = 4

    my midpoint is
     (\frac{-2+2}{2},\frac{3-1}{2})
    =(0,1)

    my distance is :

    d=\sqrt{(2+2)^2+(-1-3)^2}
    =\sqrt{32}

    What should I do next to find the center and the equation of the circle?

    thanks in advance:
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Failure's Avatar
    Joined
    Jul 2009
    From
    Zürich
    Posts
    555
    Quote Originally Posted by Anemori View Post
    I'm having hard time finding equation of a circle with given two points and a radius.

    given points (-2,3) and (2,-1) and radius = 4

    my midpoint is
     (\frac{-2+2}{2},\frac{3-1}{2})
    =(0,1)

    my distance is :

    d=\sqrt{(2+2)^2+(-1-3)^2}
    =\sqrt{32}

    What should I do next to find the center and the equation of the circle?

    thanks in advance:
    Let's assume that k:\; (x-u)^2+(y-v)^2=16 is the equation of the circle with center C(u,v) and radius 4.

    Since (-2,3) and (2,-1) lie on this circle, you get two equations for the coordinates u, v of the center, namely

    (-2-u)^2+(3-v)^2=16 and (2-u)^2+(-1-v)^2=16.
    If you now subtract the second equation from the first, you get 8u-8v+8=0, and hence u=v-1.

    Replacing u in the first equation by v-1 gives you the quadratic equation (-1-v)^2+(3-v)^2=16.
    Now solve for v: this gives you two solutions v_1,v_2 for v. Finally, pair these two solutions with the corresponding x-coordinates of C, namely u_{1,2}=v_{1,2}-1, and you are done.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jan 2010
    Posts
    142
    Quote Originally Posted by Failure View Post
    Let's assume that k:\; (x-u)^2+(y-v)^2=16 is the equation of the circle with center C(u,v) and radius 4.

    Since (-2,3) and (2,-1) lie on this circle, you get two equations for the coordinates u, v of the center, namely

    (-2-u)^2+(3-v)^2=16 and (2-u)^2+(-1-v)^2=16.
    If you now subtract the second equation from the first, you get 8u-8v+8=0, and hence u=v-1.

    Replacing u in the first equation by v-1 gives you the quadratic equation (-1-v)^2+(3-v)^2=16.
    Now solve for v: this gives you two solutions v_1,v_2 for v. Finally, pair these two solutions with the corresponding x-coordinates of C, namely u_{1,2}=v_{1,2}-1, and you are done.

    im lost where u=v-1...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member Failure's Avatar
    Joined
    Jul 2009
    From
    Zürich
    Posts
    555
    Quote Originally Posted by Anemori View Post
    im lost where u=v-1...
    Well, as I wrote, you can now replace u in (-2-u)^2+(3-v)^2=16 by v-1.

    This gives you that (-1-v)^2+(3-v)^2=16 must hold. Thus it follows from this quadratic equation that v_1=-1 or v_2=3.

    Since we know that u=v-1, we get for the corresponding x-coordinates u_{1,2} of the center of the circle, that u_1=v_1-1=-2 and u_2=v_2-1=2, respectively.

    Thus the center of the circle is either C_1(-2,-1) or C_2(2,3), and the corresponding equations of these two circles are

    k_1:\; (x+2)^2+(y+1)^2=4^2 and k_2:\; (x-2)^2+(y-3)^2=4^2.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 7
    Last Post: March 15th 2010, 04:10 PM
  2. Finding radius of circle
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: March 8th 2010, 04:02 PM
  3. Finding the radius of a circle.
    Posted in the Geometry Forum
    Replies: 1
    Last Post: April 3rd 2009, 11:00 PM
  4. finding radius of a circle - please help!
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: October 18th 2008, 11:45 AM
  5. Replies: 2
    Last Post: May 23rd 2007, 05:50 PM

Search Tags


/mathhelpforum @mathhelpforum