Answer:
x^2-4x+y^2+6y-108=0
Step-by-step explanation:
[tex]The- equation- of- circle- with -center- at- (h,k) -and -a -radius- of- r -is: \\(x-h)^2 +(y-k)^2 = r^2\\h = 2 , \\ k = -3\\r = 11\\(x-2)^2+(y-(-3))^2 = 11^2\\(x-2)^2+(y+3)^2 = 121\\x^2-4x+4 +y^2+6y+9 = 121\\x^2 -4x+y^2+6y+4+9=121\\x^2 -4x+y^2+6y+13=121\\x^2 -4x+y^2+6y=121-13\\x^2 -4x+y^2+6y= 108\\x^2 -4x+y^2+6y-108 = 0[/tex]
