Answer
$\rightarrow y^2=C^2-2Cx$
Work Step by Step
Given:
$$yy'=\sqrt x^2+y^2 -x$$
$$\frac{dy}{dx}=\frac{\sqrt x^2+y^2 -x}{y}$$
Let $y=vx \rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$
Subtituting:
$$x\frac{dv}{dx}+v=\frac{\sqrt x^2+(vx)^2 }{vx}$$
$$x\frac{dv}{dx}+v=\frac{\sqrt 1+v^2-1}{v}$$
$$x\frac{dv}{dx}=\frac{\sqrt 1+v^2-1}{v}-v$$
$$x\frac{dv}{dx}=\frac{\sqrt 1+v^2-1-v^2}{v}$$
$$\frac{dx}{x}=\frac{v}{\sqrt 1+v^2-1-v^2}dv$$
Integrating left sides:
$\int \frac{v}{\sqrt 1+v^2-1-v^2}dv$
Let $\sqrt 1+v^2=u \rightarrow \frac{v}{\sqrt 1+v^2}dv=du$
$$\int \frac{v}{\sqrt 1+v^2-1-v^2}dv$$
$$\int \frac{1}{1-\sqrt 1+v^2}\frac{v}{\sqrt 1+v^2}dv=-\ln (1-\sqrt 1+v^2)$$
Hence,
$$-\ln (1-\sqrt 1+v^2)=\ln x + \ln C$$
where $C$ is a constant of integration.
$$\ln(x(1-\sqrt 1+v^2))=\ln C$$
$$x(1-\sqrt 1+v^2)=C$$
Subtitute when $y=vx $
$$\rightarrow x-\sqrt x^2+y^2=C$$
$$\rightarrow (x-C)^2=x^2 + y^2$$
$$\rightarrow y^2=C^2-2Cx$$
