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Math Help - circle geomtery

  1. #1
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    circle geomtery

    A circle passes through the points (0 , -3) and (0, 3) and has radius 5. Find thier equations.
    how do I go about doing this, if I try to find the mid-point or distance I just end up getting zer0 ?
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  2. #2
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    Quote Originally Posted by Tweety View Post
    how do I go about doing this, if I try to find the mid-point or distance I just end up getting zer0 ?
    You are correct so far.
    So the center of the circle is on the x-axis: say (h,0).
    Now find h by noticing the distance from (h,0) to (0,3) is 5.
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  3. #3
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    Quote Originally Posted by Plato View Post
    You are correct so far.
    So the center of the circle is on the x-axis: say (h,0).
    Now find h by noticing the distance from (h,0) to (0,3) is 5.

     \sqrt{ (0-h)^2 + (3-0)^2} = 5

     \sqrt{h^2 +9} = 5

     \sqrt{h^2} + 3 = 5

     \sqrt{h^2} = 2

    does h = 2 ?

    Is my working correct?

    And how do I work out the equation from this ?


    Thanks.
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  4. #4
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    Quote Originally Posted by Tweety View Post
     \sqrt{ (0-h)^2 + (3-0)^2} = 5
     \sqrt{h^2 +9} = 5
    {\color{red} \sqrt{h^2} + 3} = 5
    Tweety, you need to learn to do basic algebra.
    The red part above is just wrong.
    <br />
\begin{gathered}<br />
  \sqrt {h^2  + 9}  = 5 \hfill \\<br />
  h^2  + 9 = 25 \hfill \\<br />
  h = ? \hfill \\ <br />
\end{gathered}
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  5. #5
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    Quote Originally Posted by Plato View Post
    Tweety, you need to learn to do basic algebra.
    The red part above is just wrong.
    <br />
\begin{gathered}<br />
\sqrt {h^2 + 9} = 5 \hfill \\<br />
h^2 + 9 = 25 \hfill \\<br />
h = ? \hfill \\ <br />
\end{gathered}
    I get it thanks
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  6. #6
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    Talking

    Quote Originally Posted by Tweety View Post
    A circle passes through the points (0 , -3) and (0, 3) and has radius 5. Find thier equations.
    Since the two points listed are not 10 units apart, they are not on opposite ends of a diameter line. You only know that, wherever the center (h, k) might be, the two points are five units away.

    So plug the two points and the (currently unknown) center into the Distance Formula (or, more usefully, the square thereof), using the fact that the two distances, by definition, must be equal to the square of the radius:

    . . . . . (h\, -\, 0)^2\, +\, (k\, +\, 3)^2\, =\, 25

    . . . . . (h\, -\, 0)^2\, +\, (k\, -\, 3)^2\, =\, 25

    These equations simplify as:

    . . . . . h^2\, +\, (k\, +\, 3)^2\, =\, 25

    . . . . . h^2\, +\, (k\, -\, 3)^2\, =\, 25

    Subtract the two equations:

    . . . . . (k\, +\, 3)^2\, -\, (k\, -\, 3)^2\, =\, 0

    . . . . . k^2\, +\, 6k\, +\, 9\, =\, k^2\, -\, 6k\, +\, 9

    . . . . . 6k\, =\, -6k

    . . . . . 12k\, =\, 0

    What then is the value of k?

    Once you have a value for k, plug this into either of the equations generated by the Distance Formula, and solve for the value of h. (You should get the same two solutions as previous profferred. This is because there are two circles which fit the requirements.)

    Once you have the center(s), plug this information, along with the radius value, into the circle equation.
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