Answer :
x^2+8x+16+y^2-4y+4=-11+16+4
[x+4]^2+[y-2]^2=9
radius is 3 and center is [-4,2]
[x+4]^2+[y-2]^2=9
radius is 3 and center is [-4,2]
[tex]x^2 + y^2+8x-4y=-11 \\ \\ x^2+8x + y^2 -4y =-11\\ \\ (x^2+8x) + ( y^2 -4y) =-11\\ \\(x^2+8x+16) + ( y^2 -4y+4)-16-4 =-11 \\ \\(x+4)^2+(y-2)^2-20=-11[/tex]
[tex](x+4)^2+(y-2)^2 =-11+20\\ \\(x+4)^2+(y-2)^2 =9 \\ \\x+4=0 \ \ and \ \ y-2 =0 \\ \\ x=-4 \ \ and \ \ y=2 \\ \\ So \ the \ center \ of \ the \ circle \ is (-4, 2)[/tex]
[tex]The \ radius \ of \ the \ circle \ is \ the \ square \ root \ of \ the \ right \\ \\ side \ of \ the \ equation \ for \ this \ circle. \\ \\ So\ the \ radius \ of \ this \ circle \ is \ \sqrt{9} \ so \ the \ radius \ of \ the \ circle \ is \ 3[/tex]
[tex](x+4)^2+(y-2)^2 =-11+20\\ \\(x+4)^2+(y-2)^2 =9 \\ \\x+4=0 \ \ and \ \ y-2 =0 \\ \\ x=-4 \ \ and \ \ y=2 \\ \\ So \ the \ center \ of \ the \ circle \ is (-4, 2)[/tex]
[tex]The \ radius \ of \ the \ circle \ is \ the \ square \ root \ of \ the \ right \\ \\ side \ of \ the \ equation \ for \ this \ circle. \\ \\ So\ the \ radius \ of \ this \ circle \ is \ \sqrt{9} \ so \ the \ radius \ of \ the \ circle \ is \ 3[/tex]
