Answer :

[tex] x^2+y^2-4x-6y+9=0\\\\ x^2-4x+16+y^2-6y+9-16=0\\\\(x-4)^2+(y-3)^2=16\\\\this\ is\ the\ equation\ of\ a\ circle:\ \ (x-a)^2+(y-b)^2=r^2\\\\center\ \ S=(4;3)\ \ \ and\ \ \ radius\ \ r =4[/tex]

Other Questions