*Right after substituting $y=vx$, Is it because we assume $x>0$?
$y\,dy + x\,dx = \sqrt{x^2 + y^2}\,dx$ $\implies y\,dy = (\sqrt{x^2 + y^2} - x)\,dx$ $\implies \dfrac{\mathrm dy}{\mathrm dx} = \dfrac{\sqrt{x^2 + y^2} - x}{y}$ $\implies \dfrac{\mathrm d}{\mathrm dx} (v x) = \dfrac{\sqrt{x^2 + v^2 x^2} - x}{v x}$
substituting $y = v x$ gives
$$\implies v + x\dfrac{\mathrm dv}{\mathrm dx} = \dfrac{\sqrt{1+ v^2} - 1}{v} \implies x\dfrac{\mathrm dv}{\mathrm dx} = \dfrac{\sqrt{1+ v^2} - 1}{v} - v$$ $$\implies x\dfrac{\mathrm dv}{\mathrm dx} = \dfrac{\sqrt{1+ v^2} - 1 - v^2}{v} \implies \dfrac{v}{\sqrt{1 + v^2} - 1 - v^2} \,dv= \dfrac{dx}{x}$$ $$\implies \dfrac{v}{\sqrt{1 + v^2} (1 - \sqrt{1 + v^2})}\,dv = \dfrac{dx}{x}$$
Let $A =\displaystyle\int \dfrac{v}{\sqrt{1 + v^2} (1 - \sqrt{1 + v^2})}\,dv $ and $1 - \sqrt{1 + v^2} = t$ $$\implies -\dfrac{1}{2} (1 + v^2)^{-\frac{1}{2}} (2 v) dv = dt \implies -\dfrac{v}{\sqrt{1 + v^2}}\,dv = dt$$ $$\implies A = -\displaystyle\int \dfrac{dt}{t} = -\ln t = - \ln (1 - \sqrt{1 + v^2})$$ $$\implies\displaystyle\int \dfrac{v}{\sqrt{1 + v^2} (1 - \sqrt{1 + v^2})}\,dv = - \ln (1 - \sqrt{1 + v^2})$$ $$\dfrac{v}{\sqrt{1 + v^2} - 1 - v^2} \,dv= \dfrac{dx}{x}$$
Integrating both sides
$$\displaystyle\int \dfrac{v}{\sqrt{1 + v^2} - 1 - v^2} \,dv = \displaystyle\int \dfrac{dx}{x} \implies -\ln (1 - \sqrt{1 + v^2}) = \ln x + \ln c$$ $$\implies (1 - \sqrt{1 + v^2})^{-1} = c x \implies 1 - \sqrt{1 + v^2} = \dfrac{1}{ c x} \implies 1 - \sqrt{1 + \dfrac{y^2}{x^2}} = \dfrac{1}{c x}$$ $$\implies 1 - \dfrac{\sqrt{x^2 + y^2}}{x} = \dfrac{1}{c x} \implies \dfrac{x - \sqrt{x^2 + y^2}}{x} = \dfrac{1}{c x}$$ $$\implies x - \sqrt{x^2 + y^2} = \dfrac{1}{c} = C \implies -\sqrt{x^2 + y^2} = C - x \implies x^2 + y^2 = (C - x)^2 = C^2 + x^2 - 2 C x \implies y^2 = C^2 - 2 C x \implies y^2 + 2 C x - C^2 = 0$$
Note that: $$y\,dy + x\,dx = \sqrt{x^2 + y^2}\,dx$$ Is simply: $$\dfrac {d(x^2+y^2) }{ \sqrt{x^2 + y^2}}=2dx$$ $$\int \dfrac {du }{ \sqrt{u}}=x^2+c_1$$ $$ \sqrt{u}=\dfrac 12x^2+C$$ And $u=x^2+y^2 >0$. $$ \sqrt{x^2+y^2}=\dfrac 12x^2+C$$