A curve amongst the family of curves represented by the differential equation, $$(x^2 - y^2)dx + 2xy \; dy = 0$$ which passes through $$(1, 1)$$, is:

We are given the differential equation

$$ (x^2-y^2)\,dx+2xy\,dy=0 $$

To see it in the familiar $$\dfrac{dy}{dx}$$ form, divide by $$dx$$:

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

Hence

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

Both the numerator $$x^2-y^2$$ and the denominator $$2xy$$ are homogeneous expressions of the same degree (degree 2). When an ODE is homogeneous, the standard substitution is

$$ y=vx \quad\text{(let }v=\dfrac{y}{x}\text{)}, $$

because it converts the equation into a separable form. First write $$y=vx$$ and then compute

$$ \dfrac{dy}{dx}=v+x\dfrac{dv}{dx} \quad\text{(using }\dfrac{d}{dx}(vx)=v+x\dfrac{dv}{dx}\text{)}. $$

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

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

Simplify the right-hand side:

$$ x^2-v^2x^2 \;=\;x^2(1-v^2), $$

and

$$ 2x(vx)=2v x^2. $$

Therefore

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

Move the first term to the right so that only $$\dfrac{dv}{dx}$$ remains on the left:

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

Put everything over a common denominator $$2v$$:

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

Thus

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

Separate the variables:

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

Now integrate both sides. We use the standard integrals

$$\int \dfrac{2v}{1+v^2}\,dv=\ln(1+v^2) \quad\text{and}\quad \int \dfrac{dx}{x}=\ln|x|.$$

Carrying out the integrations gives

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

where $$C$$ is the constant of integration. Combine the logarithms:

$$ \ln(1+v^2)+\ln|x|=C. $$

Using the property $$\ln a+\ln b=\ln(ab)$$, we have

$$ \ln\!\bigl(x(1+v^2)\bigr)=C. $$

Exponentiating removes the logarithm (i.e. $$e^{\ln(\cdot)}=\cdot$$):

$$ x(1+v^2)=K, $$

where $$K=e^{C}$$ is another (non-zero) constant. Replace $$v$$ by $$y/x$$:

$$ 1+v^2=1+\Bigl(\dfrac{y}{x}\Bigr)^2=\dfrac{x^2+y^2}{x^2}. $$

Substituting this back gives

$$ x\left(\dfrac{x^2+y^2}{x^2}\right)=K \quad\Longrightarrow\quad \dfrac{x^2+y^2}{x}=K. $$

Multiply by $$x$$ to clear the denominator:

$$ x^2+y^2=Kx. $$

This is the required family of integral curves. Rearranging puts it in a more recognisable form:

$$ x^2-Kx+y^2=0. $$

Complete the square in $$x$$. Using the identity $$(x-\tfrac{K}{2})^2=x^2-Kx+\tfrac{K^2}{4},$$ add and subtract $$\tfrac{K^2}{4}$$:

$$ \bigl(x-\tfrac{K}{2}\bigr)^2-\tfrac{K^2}{4}+y^2=0 \quad\Longrightarrow\quad \bigl(x-\tfrac{K}{2}\bigr)^2+y^2=\Bigl(\tfrac{|K|}{2}\Bigr)^2. $$

This is the equation of a circle whose centre is $$\bigl(\tfrac{K}{2},0\bigr)$$—clearly on the $$x$$-axis—and whose radius is $$\tfrac{|K|}{2}$$.

To find the particular member of the family that passes through $$(1,1)$$, substitute $$x=1,\; y=1$$ into $$x^2+y^2=Kx$$:

$$ 1^2+1^2=K\cdot1 \quad\Longrightarrow\quad K=2. $$

Putting $$K=2$$ in $$x^2+y^2=Kx$$ gives

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

Again complete the square:

$$ (x-1)^2+y^2=1. $$

This confirms that the required curve is a circle with centre $$(1,0)$$ lying on the $$x$$-axis. Therefore, among the given statements, the correct description is:

A. A circle with centre on the $$x$$-axis.

Hence, the correct answer is Option A.