# How do you write the equation for a circle in standard form? 9x^+9y^-54x-36y+116=0

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 jcaron2 Posts: 983, Reputation: 1034 Senior Member #2 Mar 21, 2011, 06:44 AM
Since the standard form looks like this:

$(x-p)^2+(y-q)^2=r^2$,

(where p and q are the x- and y-coordinates of the circle's center, and r is the radius), the first thing you need to do is divide everything by 9 so the $x^2$ and $y^2$ terms have no coefficients. Then you need to collect all the terms with x together, then do the same for the terms with why, then determine what additional constant needs to be included with each group to make it a perfect square (this process is called "completing the square").

For example, let's rewrite your equation from above:

$9x^2-54x+9y^2-36y + 116 = 0$

$x^2-6x+y^2-4y + \frac{116}{9} = 0$

$(x^2-6x + p^2) + (y^2 - 4y + q^2) = -\frac{116}{9} + p^2 + q^2$

Notice that I added unknown numbers $p^2$ and $q^2$ to both sides. Now we can figure out what each one has to be to make the two groups in parentheses perfect squares. Let's do the first one:

$x^2-6x + p^2 = (x-p)^2$

The left side of the equation is what we've got. The right side is what we want it to look like (a perfect square just like in the standard form of the equation of a circle). When we multiply out the perfect square, this is what it looks like:

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

Now we can determine what the value of p has to be to make the two expressions equal:

$x^2-6x + p^2 = x^2 - 2px + p^2$

$-6x = -2px$

$p = 3$

$p^2 = 9$

So now we have

$(x^2-6x + 9) + (y^2 - 4y + q^2) = -\frac{116}{9} + 9 + q^2$

$(x - 3)^2 + (y^2 - 4y + q^2) = -\frac{35}{9} + q^2$

Now you just have to repeat the process and complete the square for the y group to find q, and you'll be done.

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