Question 1085764
Complete the square to get the equation into the general equation of a circle.
{{{(x-h)^2+(y-k)^2=R^2}}}
I'll do one, you do the others in the same way.
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{{{8x^2+8y^2-28x+12y+221=0}}}
I'll assume you made a mistake and that it's {{{-28x}}} and not {{{-28}}}.
{{{(x^2-(7/2)x)+(y^2+(3/2))+221/28=0}}}
{{{(x^2-(7/2)x+(7/4)^2)+(y^2+(3/2)x+(3/4)^2)+221/28=(7/4)^2+(3/4)^2}}}
{{{(x-7/4)^2+(y+3/4)^2+221/28=49/16+9/16}}}
{{{(x-7/4)^2+(y+3/4)^2=58/16-221/28}}}
{{{(x-7/4)^2+(y+3/4)^2=406/112-884/112}}}
{{{(x-7/4)^2+(y+3/4)^2=-478/112}}}
Since the radius term is negative, there is no point at all. 
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If it equals zero, then the center is the single point. 
If its positive, then the square root of the value is the radius of the circle.
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Do the others the same way.