# Thread: Circles and Standard Form

1. ## 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

2. Originally Posted by aperkin2
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
$\displaystyle 15x^2+15y^2+6x-10y+\frac{3}{5}=0|:15$

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

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

$\displaystyle (x-(-\frac{1}{5}))^2+(y-\frac{1}{3})^2-(\frac{1}{3})^2=0$

3. Hello, Ashley!

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

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

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

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

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

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

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

4. Originally Posted by aperkin2
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.

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

Next, move the constant to the right side.

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

Now, group the variables.

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

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.

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

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.

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

Which translates neatly into

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

So, now what do you think?

5. Thank you so much!

6. 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!