Simmons 7.10 & 7.11: Find Curves Intersecting at Angle pi/4

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>10. Let a family of curves be integral curves of a differential equation ##y^{\prime}=f(x, y) .## Let a second family have the property that at each point ##P=(x, y)## the angle from the curve of the first family through ##P## to the curve of the second family through ##P## is ##\alpha .## Show that the curves of the second family are solutions of the differential equation
$$
y^{\prime}=\frac{f(x, y)+\tan \alpha}{1-f(x, y) \tan \alpha}
$$
>11. Use the result of the preceding problem to find the curves that form the angle ##\pi / 4## with

>(a) all straight lines through the origin;

>(b) all circles ##x^{2}+y^{2}=c^{2}##

>(Simmons, problem 7.10 and 7.11)

I was trying to work out problem 11. The solutions given are
11.(a) ##r=c e^{\theta}##
11.(b) ##r=c e^{-\theta}##

My solution:
11.
##y=mx##, ##f(x,y)=m##, ##\tan \frac{\pi}{4}=1##,thus $$\frac{dy}{dx}=\frac{m+1}{1-m}=C_1$$
##y=r \sin\theta##, ##x=r \cos\theta##, ##dy= \sin\theta dr##, ##dx=-r \sin\theta d\theta##, substituting into our differential equation,

\begin{align}
\frac{dy}{dx} &= C_1\\
dy &= C_1 dx\\
\sin\theta dr &= C_1 (-r \sin\theta d\theta)\\
\frac{1}{r} dr &= C_2 d\theta\\
\log r &= C_2 \theta\\
r &= Ce^{\theta} \blacksquare\\
\end{align}

>Edit:
I realized I made an illegal move: it should be ##\log r = C_2 \theta +C_3##, thus ##r=Ce^{C_2\theta}##, which is umm...

12.
##x^2+y^2=c^2##, ##r=c##, ##y=r \sin\theta##, ##x=r \cos\theta##
\begin{align}
\frac{dy}{dx}&=-\frac{x}{\sqrt{c^2-x^2}}\\
&=-\frac{\cos\theta}{\sin\theta}\\
\end{align}

Substituing into the equation from problem 10,

\begin{align}
\frac{dy}{dx}&=\frac{-\frac{\cos\theta}{\sin\theta}+1}{1+\frac{\cos\theta}{\sin\theta}}\\
\frac{dy}{dx}&=\frac{-\frac{\cos\theta}{\sin\theta}+1}{1+\frac{\cos\theta}{\sin\theta}}\\
dy&=\frac{-\frac{\cos\theta}{\sin\theta}+1}{1+\frac{\cos\theta}{\sin\theta}}dx\\
dr&=\frac{-\frac{\cos\theta}{\sin\theta}+1}{1+\frac{\cos\theta}{\sin\theta}}(-r)d\theta\\
\end{align}

How am I supposed to continue? Can I pull off the same trick as before? Thanks in advance.

P.S. I read this post [Express the differential equation that solves the below problem in polar form](https://math.stackexchange.com/questions/909582/express-the-differential-equation-that-solves-the-below-problem-in-polar-form)
 
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Your starting point is <br /> \frac{dy}{d\theta} = \frac{1 + f(x,y)}{1 - f(x,y)} \frac{dx}{d\theta} where the chain rule gives <br /> \begin{align*}<br /> \frac{dx}{d\theta} &amp;= \frac{dr}{d\theta} \cos \theta - r \sin \theta, \\<br /> \frac{dy}{d\theta} &amp;= \frac{dr}{d\theta}\sin \theta + r\cos \theta.<br /> \end{align*}

For part (a): The straight line through the origin which passes through (x,y) has f(x,y) = \frac{y}{x}.

For part (b): You already found f(x,y) = -\cot \theta. Proceed from here with the above expressions.
 
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