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Thread: Bit of help please - Cir. Eq.

  1. #1
    ADY
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    Bit of help please - Cir. Eq.

    Can someone check and help my methodology and confirm my answer.

    Ok, so the following equation is a circle - and im trying to find the centre and radius


    $\displaystyle x^2 - 10x + y^2 + 8y + 36 = 0$

    First Consider

    $\displaystyle x^2 - 10x$

    Compare with

    $\displaystyle x^2 + 2px = (x + p)^2 - p^2$

    we can match

    $\displaystyle x^2 - 10x$ with LHS by putting

    $\displaystyle 2p = -10$ that is $\displaystyle p = -5$

    Putting values into RHS, obtain completed square form

    $\displaystyle x^2 - 10x = (x - 10)^2 - (-10)^2 = (x + 10 (-10) - 100$

    Now

    $\displaystyle y^2 + 8y$ with y in place of x

    $\displaystyle y^2 + 2py = (y + p)^2 - p^2$

    We can match $\displaystyle y^2 + 8y$ with RHS by putting

    $\displaystyle 2p = 8$ that is $\displaystyle p = 4$

    Putting values into RHS we obtain completed square form

    $\displaystyle y^2 + 8 = (y + 4)^2 - 4^2 = (y + 4)^2 - 16$

    $\displaystyle (x - 10)^2 - 100 + (y + 4)^2 - 16 + 36 = 0$

    $\displaystyle (x - 10)^2 + (y + 4)^2 = -80$

    Hence centre (10,4) and radius -80

    Thanks for help
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  2. #2
    MHF Contributor red_dog's Avatar
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    $\displaystyle (x^2-10x+25)+(y^2+8y+16)-5=0$

    $\displaystyle (x-5)^2+(y+4)^2=5$
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  3. #3
    A riddle wrapped in an enigma
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    Hi ADY,

    Your solution (however wrong) seems to be a bit convoluted.

    Look at it simply.

    $\displaystyle x^2-10x+y^2+8y+36=0$

    First, group your x and y, and move +36 to the right side.

    $\displaystyle x^2-10x+y^2+8y=-36$

    The trick here is to add a constant to the x and y group that will make the resulting trinomial a perfect square.

    How do we do that?

    Take half of the coefficient of the linear term, square it, then add it to the group. Do that for both the x and y group. And since we added a certain amount to the left side, we must add that same amount to the right side.

    So, here we go.

    Half of -10 is -5. We square that to get 25 and we add that to the x group to get $\displaystyle x^2-10x{\color{blue}+25}$.

    Half of 8 is 4. We square that to get 16 and we add that to the y group to get $\displaystyle y^2+8x{\color{red}+16}$.

    We added a total of 25 + 16 on the left, so we add that amount to the right to keep things balanced.

    $\displaystyle (x^2-10x {\color{blue}+25})+(y^2+8x{\color{red}+16})=-36{\color{blue}+25}{\color{red}+16}$

    Finally, grouping the two perfect square trinomials, we arrive at

    $\displaystyle \boxed{(x-5)^2+(y+4)^2=5}$

    Center (5, -4), Radius = $\displaystyle \sqrt{5}$
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  4. #4
    ADY
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    Superb explaination Masters (again), thank you!
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