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Thread: One more time with Circles

  1. #1
    Member M670's Avatar
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    One more time with Circles

    Find the coordinates of the center and radius of the circle.
    $\displaystyle x^2+y^2-8x+10y=-5$
    So I like to rewrite it
    $\displaystyle x^2-8x+y^2+10y=-5$
    then I write it as the standard circle equation
    $\displaystyle (x-h)^2+(y-K)^2=r^2$

    So If I did this right I should have $\displaystyle (x-4)^2+(y+5)^2=-46$
    and then my $\displaystyle (h,k) is (4,-5) and r=\sqrt46$
    Please I need someone to confirm to me I did this properly?
    Side note $\displaystyle x^2-8x$ is expanded to $\displaystyle (x-4)^2+16$ and I move the 16 over to the other side of the = sign in the equations.
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  2. #2
    Member M670's Avatar
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    Re: One more time with Circles

    I messed this one up.... my math is wrong
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  3. #3
    MHF Contributor MarkFL's Avatar
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    Re: One more time with Circles

    You are correct to rewrite as:

    $\displaystyle x^2-8x+y^2+10y=-5$

    Now, we want to complete the square on x and y. Whatever we add to the left side, we must add to the right.

    $\displaystyle (x^2-8x+16)+(y^2+10y+25)=-5+16+25$

    Now, write everything as squares:

    $\displaystyle (x-4)^2+(y+5)^2=6^2$

    From this, we know the center is (4,-5) and the radius is 6.
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