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Thread: Prove

  1. #1
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    Prove

    show that the circle $\displaystyle x^2+y^2-2x-2y+4=0$ touches the $\displaystyle y$-axis
    I found the radius to be $\displaystyle \sqrt{1^2+1^2-4}=\sqrt{-2}$
    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 $\displaystyle x^2+y^2-2x-2y+4=0$ touches the $\displaystyle y$-axis
    I found the radius to be $\displaystyle \sqrt{1^2+1^2-4}=\sqrt{-2}$
    is there a problem with this question?
    Yes there is!

    A circle is...

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

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

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

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

    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.
    $\displaystyle y^2-2y+4=0$ 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
    $\displaystyle y^2-2y+4=0$ has no real solution.
    Thanks for pointing that out. The question seems to be wrong.
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