Results 1 to 6 of 6

Math Help - Find Center, Radius and Graph Each Circle

  1. #1
    Banned
    Joined
    Jan 2007
    Posts
    315

    Find Center, Radius and Graph Each Circle

    Find the center (h, k) and radius r of each circle below.

    Then, graph each circle.

    (1) x^2 + (y - 1)^2 = 1

    (2) x^2 + y^2 + x + y - (1/2) = 0

    (3) How do I graph such circles WITHOUT a graphing calculator?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,937
    Thanks
    338
    Awards
    1
    Quote Originally Posted by symmetry View Post
    Find the center (h, k) and radius r of each circle below.

    Then, graph each circle.

    (1) x^2 + (y - 1)^2 = 1

    (2) x^2 + y^2 + x + y - (1/2) = 0

    (3) How do I graph such circles WITHOUT a graphing calculator?
    The general equation for a circle is (x - h)^2 + (y - k)^2 = r^2 where the center of the circle is (h, k) and the circle has a radius of r.

    I won't even bother with the first one...

    2) x^2 + y^2 + x + y - \frac{1}{2} = 0

    We need to complete the square for both the x and y variables:
    (x^2 + x) +  (y^2 + y) = \frac{1}{2}

    Now, (x + a)^2 = x^2 + 2ax + a^2. So looking at the first set of parenthesis we've got x^2 + x. Equating the linear coefficient of this expression and the one right above we see that 1 = 2a, so a = \frac{1}{2}. This means we want to add a^2 = \frac{1}{4} to x^2 + x to make it a perfect square since x^2 + x + \frac{1}{4} = \left ( x + \frac{1}{2} \right ) ^2. But what we do to one side of the equation we need to do to the other. Thus:
    \left  ( x^2 + x + \frac{1}{4} \right ) +  (y^2 + y) = \frac{1}{2} + \frac{1}{4}

    \left  ( x + \frac{1}{2} \right )^2 +  (y^2 + y) = \frac{1}{2} + \frac{1}{4}

    Now to the y variable. The problem is identical and we find we need to add \frac{1}{4} to both sides again:
    \left ( x + \frac{1}{2} \right )^2 + \left ( y^2 + y + \frac{1}{4} \right ) = \frac{1}{2} + \frac{1}{4} + \frac{1}{4}

    \left  ( x + \frac{1}{2} \right )^2 + \left ( y + \frac{1}{2} \right )^2 = 1

    So this circle has a center at \left ( -\frac{1}{2}, -\frac{1}{2} \right ) and a radius of 1.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,937
    Thanks
    338
    Awards
    1
    Quote Originally Posted by symmetry View Post

    (3) How do I graph such circles WITHOUT a graphing calculator?
    Who said you had to use a graphing calculator to graph? Get some graph paper and work out some points by hand! Hey, I had to do that in High School...

    -Dan
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Banned
    Joined
    Jan 2007
    Posts
    315

    ok

    Dan,

    I follow your steps except for one: letter a in the mix.

    Where did you get the letter a?

    I got lost when you put variable a into the steps.

    What does a represent?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,937
    Thanks
    338
    Awards
    1
    Quote Originally Posted by symmetry View Post
    Dan,

    I follow your steps except for one: letter a in the mix.

    Where did you get the letter a?

    I got lost when you put variable a into the steps.

    What does a represent?
    a is just a "dummy variable." In order to complete the square for x^2 + x we need to add some constant to it. So I am comparing
    x^2 + x
    x^2 + 2ax + a^2 <-- This is the form for the perfect square (x + a)^2.

    where a^2 is the number we need to add to x^2 + x. By comparing the two expressions we can see that we need the linear x term to have the same coefficient in both lines, so 1 = 2a.

    Hopefully that explains it well enough. If not I can provide other examples.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Banned
    Joined
    Jan 2007
    Posts
    315

    ok

    I got it now.

    Thanks.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: March 12th 2011, 01:24 AM
  2. Replies: 7
    Last Post: March 15th 2010, 04:10 PM
  3. Find the center and radius of circle
    Posted in the Geometry Forum
    Replies: 3
    Last Post: December 28th 2009, 03:45 PM
  4. center and radius of a circle?
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: January 29th 2009, 09:46 PM
  5. Replies: 4
    Last Post: September 11th 2008, 09:20 AM

Search Tags


/mathhelpforum @mathhelpforum