The curve satisfying the differential equation, $$(x^2 - y^2)dx + 2xydy = 0$$ and passing through the point (1, 1) is:

We have the differential equation $$\left(x^{2}-y^{2}\right)\,dx+2xy\,dy=0.$$

First divide by $$dx$$ so that the equation is written in its derivative form:

$$\left(x^{2}-y^{2}\right)+2xy\,\dfrac{dy}{dx}=0.$$

Now isolate the derivative:

$$2xy\,\dfrac{dy}{dx}=-(x^{2}-y^{2})$$

and then

$$\dfrac{dy}{dx}=-\dfrac{x^{2}-y^{2}}{2xy}.$$

Because every term is a power of $$x$$ and $$y$$ of the same total degree, the right-hand side is a homogeneous function. Hence we use the standard substitution $$y=vx,$$ where $$v$$ is a function of $$x$$.

With $$y=vx$$ we have $$\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}.$$ Substituting both $$y=vx$$ and $$\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$$ into the differential equation gives

$$v+x\dfrac{dv}{dx}=-\dfrac{x^{2}-(vx)^{2}}{2x(vx)}.$$

Simplifying the numerator and denominator on the right:

$$v+x\dfrac{dv}{dx}=-\dfrac{x^{2}(1-v^{2})}{2x^{2}v}=-\dfrac{1-v^{2}}{2v}.$$

Now shift the first term to isolate the derivative again:

$$x\dfrac{dv}{dx}=-\dfrac{1-v^{2}}{2v}-v=\dfrac{-(1-v^{2})-2v^{2}}{2v}=-\dfrac{1+v^{2}}{2v}.$$

Therefore

$$\dfrac{dv}{dx}=-\dfrac{1+v^{2}}{2vx}.$$

Separate the variables:

$$\dfrac{2v}{1+v^{2}}\,dv=-\dfrac{dx}{x}.$$

The integral formula $$\displaystyle\int \dfrac{2v}{1+v^{2}}\,dv=\ln(1+v^{2})$$ is now used. Integrating both sides we obtain

$$\ln(1+v^{2})=-\ln x + C,$$

where $$C$$ is the constant of integration. Exponentiating gives

$$1+v^{2}=C_{1}\dfrac{1}{x}, \quad\text{where }C_{1}=e^{C}.$$\

Replace $$v$$ by $$y/x$$ (because $$v=y/x$$):

$$1+\left(\dfrac{y}{x}\right)^{2}=\dfrac{C_{1}}{x}.$$

Multiply every term by $$x^{2}$$ to clear denominators:

$$x^{2}+y^{2}=C_{1}x.$$

Let $$C_{1}=k$$ for simplicity. Then

$$x^{2}+y^{2}=kx.$$

To decide the constant $$k$$ we employ the given point $$(1,1).$$ Substituting $$x=1,\;y=1$$ gives

$$1^{2}+1^{2}=k\cdot1 \;\Longrightarrow\; 2=k.$$

Hence $$k=2$$ and the explicit equation of the curve is

$$x^{2}+y^{2}=2x.$$

Complete the square for the $$x$$-terms to identify the curve:

$$x^{2}-2x+y^{2}=0 \;\Longrightarrow\; (x-1)^{2}+y^{2}=1^{2}.$$

This is a circle with centre $$(1,0)$$ and radius $$1.$$

Hence, the correct answer is Option B.