Consider a triangle $$\triangle$$ whose two sides lie on the x-axis and the line $$x + y + 1 = 0$$. If the orthocenter of $$\triangle$$ is (1,1), then the equation of the circle passing through the vertices of the triangle $$\triangle$$ is

Let the two given lines be
    $$y = 0$$  (the x-axis) and  $$x + y + 1 = 0 \;\Longrightarrow\; y = -x - 1$$.

Since the two sides of the triangle lie on these lines, their intersection becomes a vertex of the triangle.
Intersection of the two lines: put $$y = 0$$ in $$x + y + 1 = 0$$ to get $$x = -1$$.
Thus one vertex is $$P(-1,0)$$.

Let the remaining vertices be
    $$A(a,0)$$ on the x-axis and $$B(b,-b-1)$$ on $$x + y + 1 = 0$$.

The orthocenter is given as $$H(1,1)$$. Altitudes pass through the orthocenter, so we use this fact for all three vertices.

Altitude from $$A$$:
Side $$PB$$ lies on $$y = -x - 1$$ having slope $$m_{PB} = -1$$.
Hence the altitude from $$A$$ is perpendicular to it and has slope $$+1$$.
Line through $$A(a,0)$$ of slope 1 is $$y = x - a$$.
Since this line must pass through $$H(1,1)$$: $$1 = 1 - a \Longrightarrow a = 0$$.
Therefore $$A(0,0)$$.

Altitude from $$B$$:
Side $$PA$$ lies on the x-axis, slope $$0$$, so its perpendicular altitude is a vertical line.
Thus the altitude from $$B$$ is $$x = \text{constant}$$, and it must go through $$H(1,1)$$, giving $$x = 1$$.
Hence $$b = 1$$ and $$B(1,-2)$$ because $$y = -b - 1 = -2.$$

We have now fixed all three vertices:
    $$P(-1,0),\; A(0,0),\; B(1,-2).$$

The circumcircle through these points is taken as $$x^{2}+y^{2}+gx+fy+c=0.$$ Substitute each vertex:

For $$A(0,0)$$:  $$0+0+0+0+c=0 \;\Longrightarrow\; c=0.$$

For $$P(-1,0)$$:  $$1 + 0 - g + 0 + 0 = 0 \;\Longrightarrow\; g = 1.$$

For $$B(1,-2)$$:  $$1 + 4 + g(1) + f(-2) + 0 = 0.$$ With $$g = 1$$ this becomes $$6 - 2f = 0 \;\Longrightarrow\; f = 3.$$

Therefore the required circle is
$$x^{2} + y^{2} + x + 3y = 0.$$

Hence, the correct option is:
Option B which is: $$x^2 + y^2 + x + 3y = 0$$.