1. Given differential equation

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

   We need the curve passing through $(1,1)$.

2. Rewrite in differential form

   Divide by $dx$:

   $x^2-y^2 + 2xy\frac{dy}{dx}=0$

   So,

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

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

3. Use substitution to simplify

   Since the equation is homogeneous, let

   $y=vx \Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$

   Substitute into the equation:

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

   Hence,

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

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

4. Separate variables

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

   Integrate both sides:

   $\int \frac{2v}{v^2+1}\,dv = -\int \frac{dx}{x}$

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

5. Substitute back $v=\frac{y}{x}$

   $\ln\left(1+\frac{y^2}{x^2}\right) = -\ln|x| + C$

   Combine logarithms:

   $\ln\left(\frac{x^2+y^2}{x^2}\right)= -\ln|x| + C$

   This simplifies to

   $\ln(x^2+y^2)=C'$

   Therefore,

   $x^2+y^2 = C$

6. Use the point $(1,1)$

   $1^2+1^2=2$

   So,

   $x^2+y^2=2$

7. Interpret the curve

   $x^2+y^2=2$

   is a circle centered at the origin with radius

   $r=\sqrt{2}$

8. Check options

   • A: circle of radius one $d7$
   • B: hyperbola $d7$
   • C: ellipse $d7$ (though a circle is a special ellipse, in MCQ the specific description is expected)
   • D: circle of radius two $d7$

So the actual curve is a circle of radius $\sqrt{2}$, which is not listed exactly among the options.