# How do you write 2x² + 2y² - 6x + 4y + 1 = 0 in standard form?

Oct 26, 2016

Standard circle form: ${\left(x - \frac{3}{2}\right)}^{2} + {\left(y + 1\right)}^{2} = {\left(\frac{\sqrt{11}}{2}\right)}^{2}$

#### Explanation:

Note that the given equation is in standard polynomial form
so I have assumed you need it converted to standard circle form.

$2 {x}^{2} + 2 {y}^{2} - 6 x + 4 y + 1 = 0$

$\rightarrow \left(2 {x}^{2} - 6 x \textcolor{w h i t e}{\text{XXXX"))+(2y^2+4ycolor(white)("XXXX}}\right) = - 1$

$\rightarrow 2 \left({x}^{2} - 3 x \textcolor{red}{+ {\left(\frac{3}{2}\right)}^{2}}\right) + 2 \left({y}^{2} + 4 y \textcolor{b l u e}{+ 1}\right) = - 1 \textcolor{red}{+ 2 {\left(\frac{3}{2}\right)}^{2}} \textcolor{b l u e}{+ 2 \left(1\right)}$

$\rightarrow 2 {\left(x - \frac{3}{2}\right)}^{2} + 2 {\left(y + 1\right)}^{2} = \frac{11}{2}$

$\rightarrow {\left(x - \frac{3}{2}\right)}^{2} + {\left(y + 1\right)}^{2} = \frac{11}{4}$

$\rightarrow {\left(x - \frac{3}{2}\right)}^{2} + {\left(y + 1\right)}^{2} = {\left(\frac{\sqrt{11}}{2}\right)}^{2}$
which is standard circle form for a circle with center $\left(\frac{3}{2} , - 1\right)$ and radius $\frac{\sqrt{11}}{2}$