we know that
the standard form of the equation of the circle is
[tex](x-h) ^{2} +(y-k) ^{2} =r ^{2} [/tex]
where
(h,k) is the center
r is the radius
case a)
[tex] x^{2} + y^{2} -8x-8y+23=0 \\ (x-4) ^{2} +(y-4)^{2} = 3^{2} [/tex]
Is a circle
with radius r=[tex]3[/tex] units
and center [tex](4,4)[/tex]
case b)
[tex] x^{2} + y^{2} -8x-8y+32=0 \\ (x-4) ^{2} +(y-4) ^{2} =0[/tex]
Is not a circle
case c)
[tex] x^{2} + y^{2} -4x-y+23=0 \\ (x+2) ^{2} +(y-2) ^{2} =-15[/tex]
Is not a circle
case d)
[tex] x^{2} + y^{2} +4x+4y+9=0 \\ (x+2) ^{2} +(y+2) ^{2} =-1[/tex]
Is not a circle
therefore
the answer is
[tex]x^{2} + y^{2} -8x-8y+23=0[/tex]
