# How do you write the equation 4x^2+4y^2-24y+36=0 in standard form and find the center and radius?

Jun 11, 2017

Given $4 {x}^{2} + 4 {y}^{2} - 24 y + 36 = 0 \text{ }$

The a conic section of the general form:

$A {x}^{2} + B x y + C {y}^{2} + D x + E y + F = 0 \text{ }$

Matching coefficients between equations  and  we observe that:

A = B = 4

This identifies the given conic section as a circle of the general form:

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

Begin progress to the form of equation  by dividing equation  by 4:

${x}^{2} + {y}^{2} - 6 y + 9 = 0 \text{ [1.1]}$

Subtract 9 from both sides:

${x}^{2} + {y}^{2} - 6 y = - 9 \text{ [1.2]}$

Because there is no x term, we conclude that $h = 0$:

${\left(x - 0\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2} \text{ [3.1]}$

There is a y term, therefore, we add k^2 to both sides of equation [1.2]:

${x}^{2} + {y}^{2} - 6 y + {k}^{2} = {k}^{2} - 9 \text{ [1.3]}$

From the pattern ${\left(y - k\right)}^{2} = {y}^{2} - 2 k y + {k}^{2}$, we know that we can can find the value of k by setting middle term of the parttern equal to the y term:

$- 2 k y = - 6 y$

$k = 3 \leftarrow$ substitute this into equations [3.1] and [1.3]:

${\left(x - 0\right)}^{2} + {\left(y - 3\right)}^{2} = {r}^{2} \text{ [3.2]}$

${x}^{2} + {y}^{2} - 6 y + {3}^{2} = {3}^{2} - 9$

Simplify:

x^2+y^2-6y + 9 = 0 [1.4]"#

Because the right side of equation [1.4] this indicates that the radius is 0:

${\left(x - 0\right)}^{2} + {\left(y - 3\right)}^{2} = 0 \text{ [3.3]}$

This degenerates into the point $\left(0 , 3\right)$