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Math Help - Prove

  1. #1
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    Prove

    show that the circle x^2+y^2-2x-2y+4=0 touches the y-axis
    I found the radius to be \sqrt{1^2+1^2-4}=\sqrt{-2}
    is there a problem with this question?
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  2. #2
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    Quote Originally Posted by Punch View Post
    show that the circle x^2+y^2-2x-2y+4=0 touches the y-axis
    I found the radius to be \sqrt{1^2+1^2-4}=\sqrt{-2}
    is there a problem with this question?
    Yes there is!

    A circle is...

    \left(x-x_c\right)^2+\left(y-y_c\right)^2=r^2

    x^2-2x_cx+\left(x_c\right)^2+y^2-2y_cy+\left(y_c\right)^2=r^2

    From your equation.... \left(x_c,y_c)=(1,1)

    hence x^2-y^2-2x_cx-2y_cy+2-r^2=0\Rightarrow

    Your constant has to be <2 to have any radius at all for the circle centred at (1,1)
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  3. #3
    Member Traveller's Avatar
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    If the circle touches the y-axis at some point, then for that point x = 0. Substitute this value of x in the given equation and solve for y to get a solution in reals.
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  4. #4
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    Quote Originally Posted by Traveller View Post
    If the circle touches the y-axis at some point, then for that point x = 0. Substitute this value of x in the given equation and solve for y to get a solution in reals.
    y^2-2y+4=0 has no real solution.
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  5. #5
    Member Traveller's Avatar
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    Quote Originally Posted by Archie Meade View Post
    y^2-2y+4=0 has no real solution.
    Thanks for pointing that out. The question seems to be wrong.
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