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

Option B

center (3,2)

radius 3

Step-by-step explanation:

Given:

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

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

Now completing square of x^2-6x by introducing +9 on both sides:

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

(x-3)^2+y^2+4y=5

Now completing square of y^2+4y by introducing +4 on both sides:

(x-3)^2+y^2+4y+4=5+4

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

Now comparing with the circle equation:

(x-h)^2 + (y-k)^2= r^2

where

r= radius of circle

h= x-offset from origin

k= y-offset from origin

In given case

r=3

h=3

k=2

Hence, option B is correct with radius =3 and center =(3,2)!

step-by-step explanation:

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