Answer
A circle, centered at $(7,-3)$, with radius $1.$
Work Step by Step
$2x^{2}+2y^{2}-28x+12y=-114\qquad $
Divide with 2 and group like terms:
$(x^{2}-14x)+(y^{2}+6y)=-57\qquad$
Complete the square:
$(x^{2}-14x+49)+(y^{2}+6y+9)=-57+49+9$
$(x-7)^{2}+(y+3)^{2}=1$
This equation is of a circle, centered at $(7,-3)$, with radius $1.$
