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(A) The given differential equation is $(x^2 - y^2)dx + 2xy\, dy = 0$.This can be rewritten as $x^2 dx - y^2 dx + 2xy\, dy = 0$.Rearranging the terms,

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a circle with centre on the $x-$ axis

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(A) The given differential equation is $(x^2 - y^2)dx + 2xy\, dy = 0$.This can be rewritten as $x^2 dx - y^2 dx + 2xy\, dy = 0$.Rearranging the terms,we get $x^2 dx + (2xy\, dy - y^2 dx) = 0$.Dividing by $x^2$,we have $dx + \frac{2xy\, dy - y^2 dx}{x^2} = 0$.Recognizing the exact differential,this is $d(x) + d\left(\frac{y^2}{x}\right) = 0$.Integrating both sides,we get $x + \frac{y^2}{x} = C$,which simplifies to $x^2 + y^2 = Cx$.Since the curve passes through $(1, 1)$,we substitute $x=1$ and $y=1$ to find $C$: $1^2 + 1^2 = C(1) \implies C = 2$.Thus,the equation of the curve is $x^2 + y^2 = 2x$,or $(x-1)^2 + y^2 = 1$.This represents a circle with center at $(1, 0)$,which lies on the $x-$ axis.

The 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

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