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Math Help - Circles and Standard Form

  1. #1
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    Post Circles and Standard Form

    Hi! I am stuck on my homework and can't find an example in my book that looks like this problem. The directions are to put in standard form; state whther it is no graph, point graph, or a circle graph, and sketch those that have a graph. My equation is 15x^2+ 15y^2+ 6x+ 3/5 = 10y

    I don't have a clue how to even do the first step to put into standard form. This doesn't look like any of the other problems I have seen. I appreciate any help at all!

    Thanks so much!
    Ashley
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  2. #2
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    Quote Originally Posted by aperkin2 View Post
    Hi! I am stuck on my homework and can't find an example in my book that looks like this problem. The directions are to put in standard form; state whther it is no graph, point graph, or a circle graph, and sketch those that have a graph. My equation is 15x^2+ 15y^2+ 6x+ 3/5 = 10y

    I don't have a clue how to even do the first step to put into standard form. This doesn't look like any of the other problems I have seen. I appreciate any help at all!

    Thanks so much!
    Ashley
    15x^2+15y^2+6x-10y+\frac{3}{5}=0|:15

    x^2+y^2+\frac{6}{15}x-\frac{10}{15}y+\frac{3}{75}=0

    x^2+2\frac{3}{15}x+\frac{9}{225}+y^2-2\frac{5}{15}+\frac{25}{225}+\frac{9}{225}-\frac{9}{225}-\frac{25}{225}=0

    (x-(-\frac{1}{5}))^2+(y-\frac{1}{3})^2-(\frac{1}{3})^2=0
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  3. #3
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    Hello, Ashley!

    I must assume that you've done problems that require completing the square.


    15x^2+ 15y^2+ 6x+ \tfrac{3}{5} \:= \:10y
    Get all variables on the left, constants on the right:

    . . 15x^2 + 6x + 15y^2 - 10y \;=\;-\tfrac{3}{5}


    Divide by 15: . x^2 + \tfrac{2}{5}x + y^2 - \tfrac{2}{3}y \;=\;-\tfrac{3}{75}


    Complete the square: . x^2 + \tfrac{2}{5}x + {\color{blue}\tfrac{1}{25}} + y^2 - \tfrac{2}{3}y + {\color{red}\tfrac{1}{9}} \;=\;-\tfrac{3}{75} + {\color{blue}\tfrac{1}{25}} + {\color{red}\tfrac{1}{9}}


    Simplify: . \left(x + \tfrac{1}{5}\right)^2 + \left(y - \tfrac{1}{3}\right)^2 \;=\;\tfrac{1}{9}


    This circle has center \left(-\frac{1}{5},\:\frac{1}{3}\right) and radius r = \frac{1}{3}

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  4. #4
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    Quote Originally Posted by aperkin2 View Post
    Hi! I am stuck on my homework and can't find an example in my book that looks like this problem. The directions are to put in standard form; state whther it is no graph, point graph, or a circle graph, and sketch those that have a graph. My equation is 15x^2+ 15y^2+ 6x+ 3/5 = 10y

    I don't have a clue how to even do the first step to put into standard form. This doesn't look like any of the other problems I have seen. I appreciate any help at all!

    Thanks so much!
    Ashley
    Edit: Oh, well. Now you have 3 versions.

    Hi aperkin2,

    The first thing to do would be to set everything equal to zero.

    15x^2+15y^2+6x-10y+\frac{3}{5}=0

    Next, move the constant to the right side.

    15x^2+15y^2+6x-10y=-\frac{3}{5}

    Now, group the variables.

    15x^2+6x+15y^2-10y=-\frac{3}{5}

    Next, we need to complete the square for the two groups. To do this, we'll have to factor out the 15 from each group in order to make our quadratic terms have a coefficient of 1.

    15\left(x^2+\frac{2}{5}\right)+15\left(y^2-\frac{2}{3}\right)=-\frac{3}{5}

    Now, if you remember how to complete the square, and I'm sure you do, we'll end up with the following. I'll spare you the details.

    15\left(x^2+\frac{2}{5}x+\left(\frac{1}{5}\right)^  2\right)+15\left(y^2-\frac{2}{3}+\left(\frac{1}{3}\right)^2\right)=-\frac{3}{5}+\frac{3}{5}+\frac{5}{3}

    Which translates neatly into

    \left(x+\frac{1}{5}\right)^2+\left(y-\frac{1}{3}\right)^2=\frac{1}{9}

    So, now what do you think?
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  5. #5
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    Thank you so much!
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  6. #6
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    Thank you all so much! You guys seriously need to write textbooks because nobody has ever explained things to me that well before
    I have to present this problem to my class today and couldn't have done it without all of your help!
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