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Math Help - Find the points where two circles cross

  1. #1
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    Find the points where two circles cross

    I have two circles with equations:
    (x - 1)^2 + (y - 4)^2 = 25 and (x - 9)^2 + (y - 4)^2 = 16

    And I need to work out where the "lines" of the two circles cross. I can do this fine with a straight line and a circle but how do I do it with two circle equations?

    I tried (y-4)^2 same so
    (y - 4)^2 = 25 - (x - 1)^2
    (y - 4)^2 = 16 - (x - 9)^2

    and so 25 - (x - 1)^2 = 16 - (x - 9)^2

    But could not seem to get any sensible answer. Can anyone please help me out?
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  2. #2
    MHF Contributor Unknown008's Avatar
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    Expand this and solve the quadratic with x.

    9 + (x^2 - 18x + 81) = (x^2 -2x + 1)

    -18x + 90 = -2x +1
    16x = 89

    x = 89/16

    Now that you've got the x value, find the y values.
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  3. #3
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    But the circles cross in two places.
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  4. #4
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    Quote Originally Posted by angypangy View Post
    But the circles cross in two places.
    Yes, but since the 2 circle centres lie on the line y=4,
    the line y = 4 acts as an axis of symmetry for the graph.
    Therefore the common chord going through the points of contact is vertical.
    The value of x found will locate 2 y co-ordinates.
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  5. #5
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    Hello, angypangy!

    Find the intersections of: . \begin{Bmatrix} (x - 1)^2 + (y - 4)^2 &=& 25 & [1] \\ (x - 9)^2 + (y - 4)^2 &=& 16 & [2]\end{Bmatrix}

    Subtract [1] - [2]: . (x-1)^2 - (x-9)^2 \;=\;0

    . . . . (x^2 - 2x+1) - (x^2 - 18x + 81) \;=\;9

    . . . . . . x^2 - 2x + 1 - x^2 + 18x - 81 \;=\;9

    . . 16x \:=\;89 \quad\Rightarrow\quad x \:=\:\frac{89}{16}


    Substitute into [1]: . \left(\frac{89}{16}-1)^2 + (y-4)^2 \;=\;25

    . . . . . . . . . . . . . . . . . (\frac{73}{16})^2 + (y-4)^2 \;=\;25

    . . . . . . . . . . . . . . . . . . . . . . . (y-4)^2 \;=\;25-(\frac{73}{16})^2 \;=\; \frac{1071}{16^2}

    . . . . . . . . . . . . . . . . . . . . . . . . y - 4 \;=\;\pm\frac{\sqrt{1071}}{16}

    . . . . . . . . . . . . . . . . . . . . . . . . . . . y \;=\;4 \pm\frac{\sqrt{1071}}{16}


    The intersections are: . \left(\dfrac{89}{16},\;4 \pm \dfrac{\sqrt{1071}}{16} \right)

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  6. #6
    MHF Contributor Unknown008's Avatar
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    Yes, just like a line can cross a circle at two points, in this problem, the circles cut each other at two points.

    As Soroban showed you, the two coordinates are

    \left( \dfrac{89}{16}, 4+\dfrac{\sqrt{1071}}{16}\right)

    and

    \left( \dfrac{89}{16}, 4-\dfrac{\sqrt{1071}}{16}\right)
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