# How do you write the equation of the circle in standard form, identify the center and radius of 4x^2+4y^2+12x-24y+41=0?

Jan 23, 2017

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

#### Explanation:

The standard form of equation of a circle is the center-radius form of its equation and it is in the format

${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2}$, whose center is $\left(h , k\right)$ and radius is $r$.

To convert $4 {x}^{2} + 4 {y}^{2} + 12 x - 24 y + 41 = 0$ into standard form, we should group $x$-terms and $y$-terms separately as follows:

$4 {x}^{2} + 12 x + 4 {y}^{2} - 24 y = - 41$ and now competing squares, this becomes

$\left({\left(2 x\right)}^{2} + 2 \times 3 \times 2 x + {3}^{2}\right) + \left({\left(2 y\right)}^{2} - 2 \times 6 \times 2 y + {6}^{2}\right) = - 30 + {3}^{2} + {6}^{2}$

or ${\left(2 x + 3\right)}^{2} + {\left(2 y - 6\right)}^{2} = - 41 + 9 + 36$

i.e. ${\left(2 x + 3\right)}^{2} + {\left(2 y - 6\right)}^{2} = 4$ and dividing by ${2}^{2} = 4$

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

Hence, center is $\left(- \frac{3}{2} , 3\right)$ and radius is $1$
graph{4x^2+4y^2+12x-24y+41=0 [-4.125, 0.875, 1.66, 4.16]}