# Question #2d0c0

May 21, 2015

The first thing to do is get ${x}^{2}$ and ${y}^{2}$ by themselves without a coefficient. Luckily, all of the terms in this equation are multiples of 4, so I can simplify this by dividing by 4:

${x}^{2} + {y}^{2} + 4 x - 3 y = 0$

Now I need to group the like terms and add numbers to complete the squares.

$\left({x}^{2} + 4 x\right) + \left({y}^{2} - 3 x\right) = 0$

Just a quick trick - when you have the ${x}^{2}$ and the $x$ term and you need to find the 'unit' term, just take half of the $x$ term and square it.

$\left({x}^{2} + 4 x + {2}^{2}\right) + \left({y}^{2} - 3 x + {\left(- 1.5\right)}^{2}\right) = {2}^{2} + {\left(- 1.5\right)}^{2}$

$\left({x}^{2} + 4 x + 4\right) + \left({y}^{2} - 3 x + 2.25\right) = 4 + 2.25$

Now I can "unsquare" the polynomials.

${\left(x + 2\right)}^{2} + {\left(y - 1.5\right)}^{2} = 6.25$

And since we know a circle with center (a, b) and radius 'r' has the equation ${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$, we can say that this circle has a center of (-2, 1.5), and a radius of $\sqrt{6.25}$, or 2.5.