Finding a family of curves cutting the family $x^2+y^2=C$ in an angle of $\pi/4$

Question

I need to find a family of curves cutting the family $x^2+y^2=C$ in an angle of $\pi/4$.

What I did:

First, I found the relation between the slopes of the two families. Since $\theta=\arctan(\frac{y_1'-y_2'}{1+y_1'y_2'})$, we need $\frac{y_1'-y_2'}{1+y_1'y_2'}=1$, or $y_1'=\frac{1+y_2'}{1-y_2'}$.

Then I found the ODE that describes the family $x^2+y^2=C$. I got $x=-yy'$.

Now, I replaced $y'$ by $\frac{1+y'}{1-y'}$ to get an ODE that describes the family I need to find.

I got $y=\frac{y'-1}{y'+1}x$.

My question is whether what I did until now is correct, and how to solve the ODE I got to.

Answer 1

Yes,it is correct.To solve the ODE: Write $ y'=dy/dx $ and then adjust to get:
$x(dy-dx)=y(dy+dx) \implies xdy-ydx=(1/2)d(x^2+y^2) \implies x^2 (\frac{xdy-ydx}{x^2})=(1/2)d(x^2+y^2)\implies x^2d(y/x)=(1/2)d(x^2+y^2) \implies \frac{d(y/x)}{1+(y/x)^2}=(1/2)\frac{d(x^2+y^2)}{x^2+y^2}$

Integrate to get:
$\arctan(y/x)=(1/2)ln(x^2+y^2)+c$
etc...