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

##### 2 Answers

The center is

#### Explanation:

The standard equation of a circle is

Where the center is

Here, we have

Transform this equation to the standar form by completing the square

Factorise

The center is

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

Center is at

#### Explanation:

Given:

Complete the square by adding

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

where:

#(a,b)# are the coordinates of the circle's center

#r# is the radius of the circle

So, here the circle has a radius of

Graph of the circle:

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