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

Question

17278 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-04T14:29:21

The center is #=(3,-2)# and the radius is #=3#

The standard equation of a circle is

#(x-a)^2+(y-b)^2=r^2#

Where the center is #=(a,b)# and the radius is #=r#

Here, we have

#x^2+y^2-6x+4y+4=0#

Transform this equation to the standar form by completing the square

#x^2-6x+y^2+4y=-4#

#x^2-6y+9+y^2+4y+4=-4+9+4#

Factorise

#(x-3)^2+(y+2)^2=3^2#

The center is #=(3,-2)# and the radius is #=3#

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

Answer 2 · 2018-07-05T15:38:22

Center is at #(3,-2)#, radius is #3# units

Given: #x^2+y^2-6x+4y+4=0#.

#=>x^2-6x+y^2+4y+4=0#

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

#=>x^2-6x+9+y^2+4y+4=9#

#=>(x-3)^2+(y+2)^2=9#

#=>(x-3)^2+(y+2)^2=3^2#

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

#(x-a)^2+(y-b)^2=r^2#

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 #3# units, and has its center located at #(3,-2)#.

Graph of the circle:

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