Question #c5785

1 Answer
Dec 1, 2017

Please see below.

Explanation:

.

#x^2+y^2-6x-6y+9=0#

#x^2-6x+9+y^2-6y+9-9=0#

#(x-3)^2+(y-3)^2-9=0#

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

If you compare this with the standard equation of a circle:

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

where #h# and #k# are the #x# and #y# of the center of the circle and #r# is the radius, we see:

#h=3#, #k=3#, and #r=3#

which means this is circle with center at #(3,3)# with radius of #3#. It touches both #x# and #y# axes.