Answer
The graph is not a circle.
Work Step by Step
RECALL:
(1) The center-radius form of a circle's equation is $(x-h)^2+(y-k)^2=r^2$ with center at $(h, k)$ and a radius of $r$ units.
(2) The radius of a circle must be positive.
(3) To complete the square for $x^2+bx$, $(\frac{b}{2})^2$ must be added. After completing the square, the factored form of the perfect square trinomial is $(x+b)^2$.
Subtract $14$ to both sides of the equation to obtain:
$x^2+y^2+2x-6y=-14$
Group terms with the same variables to obtain:
$(x^2+2x)+(y^2-6y)=-14$
Complete the square for each binomial using the formula in (3) above.
Make sure to add to the right side of the equation whatever is added on the left side to maintain the equality of both sides.
$(x^2+2x+\color{red}{(\frac{2}{2})^2}) + (y^2-6y+\color{red}{(\frac{-6}{2})^2}) = -14 +\color{red}{(\frac{2}{2})^2+(\frac{-6}{2})^2}
\\(x^2+2x+1)+(y^2-6y+9)=-14+1+9
\\(x^2+2x+1)+(y^2-6y+9)=-4
\\(x+1)^2+(y-3)^2=-(2^2)$
This equation is in the same form as the formula/equation in (1) above. However, since the radius of a circle is positive, the square of the radius ($r^2$) should also be positive.
Since the value of $r^2$ in the equation above is negative, then the equation's graph is not a circle.
