# What are the center and radius of the circle defined by the equation x^2+y^2-6x+4y+4=0?

Jul 4, 2018

The center is $= \left(3 , - 2\right)$ and the radius is $= 3$

#### Explanation:

The standard equation of a circle is

${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$

Where the center is $= \left(a , b\right)$ and the radius is $= r$

Here, we have

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

Transform this equation to the standar form by completing the square

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

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

Factorise

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

The center is $= \left(3 , - 2\right)$ and the radius is $= 3$

graph{(x^2+y^2-6x+4y+4)=0 [-14.08, 8.11, -5.51, 5.58]}

Jul 5, 2018

Center is at $\left(3 , - 2\right)$, radius is $3$ units

#### Explanation:

Given: ${x}^{2} + {y}^{2} - 6 x + 4 y + 4 = 0$.

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

Complete the square by adding $9$ to both sides.

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

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

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

Now, compare it with the equation of a circle, given by:

${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$

where:

• $\left(a , b\right)$ are the coordinates of the circle's center

• $r$ is the radius of the circle

So, here the circle has a radius of $3$ units, and has its center located at $\left(3 , - 2\right)$.

Graph of the circle:

graph{(x-3)^2+(y+2)^2=9 [-10, 10, -5, 5]}