1. Given differential equation

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

   We need the member of this family that passes through $(1,1)$.

2. Rewrite as a differential equation in $y$ and $x$

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

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

   Instead of solving directly, observe that the differential form may be exact.

3. Check exactness

   Let $M(x,y)=x^2-y^2, \qquad N(x,y)=2xy$

   Then $\frac{\partial M}{\partial y}=-2y, \qquad \frac{\partial N}{\partial x}=2y$

   These are not equal, so it is not exact.

4. Look for a useful substitution

   Since terms are homogeneous of degree 2, write $y=vx \quad \Rightarrow \quad \frac{dy}{dx}=v+x\frac{dv}{dx}$

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

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

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

5. Integrate

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

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

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

   Hence, $x(v^2+1)=C_1$

   Since $v=\frac{y}{x}$, $x\left(1+\frac{y^2}{x^2}\right)=C_1$

   $x + \frac{y^2}{x}=C_1$

   Multiplying by $x$, $x^2+y^2=C_1x$

   Let $C_1=c$. Then family is $x^2+y^2=cx$

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

   Substitute $(x,y)=(1,1)$: $1^2+1^2=c(1)$ $2=c$

   So the required curve is $x^2+y^2=2x$

7. Identify the curve

   Rewrite: $x^2-2x+y^2=0$ $(x-1)^2 + y^2 = 1$

   This is a circle with centre $(1,0)$, which lies on the x-axis.

8. Evaluate options

   • A: circle with centre on the y-axis — false
   • B: ellipse with major axis along the y-axis — false
   • C: circle with centre on the x-axis — true
   • D: hyperbola with transverse axis along the x-axis — false

Therefore, the correct answer is C.