How do you write the equation of the circle in standard form, identify the center and radius of 9x^2+9y^2+54x-36y+17=0?

Feb 26, 2017

Equation of the circle in standard form is ${\left(x - \left(- 3\right)\right)}^{2} + {\left(y - 2\right)}^{2} = {\left(\frac{10}{3}\right)}^{2}$ and its center is $\left(- 3 , 2\right)$ and radius is $\frac{10}{3}$

Explanation:

Standard form of equation of a circle is

${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$,

which has a radius $r$ and its center is $\left(h , k\right)$

Now $9 {x}^{2} + 9 {y}^{2} + 54 x - 36 y + 17 = 0$ can be written as (dividing each term by $9$

${x}^{2} + {y}^{2} + 6 x - 4 y = - \frac{17}{9}$

or ${x}^{2} + 6 x + 9 + {y}^{2} - 4 y + 4 = - \frac{17}{9} + 9 + 4$

or ${\left(x + 3\right)}^{2} + {\left(y - 2\right)}^{2} = \frac{100}{9}$

or ${\left(x - \left(- 3\right)\right)}^{2} + {\left(y - 2\right)}^{2} = {\left(\frac{10}{3}\right)}^{2}$

and as is observed. its center is $\left(- 3 , 2\right)$ and radius is $\frac{10}{3}$
graph{9x^2+9y^2+54x-36y+17=0 [-13.125, 6.875, -3.32, 6.68]}