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Everything is wrong here. Note that if you take a circle, say, $r=2$, then $dr/d\theta=0$. And yet the slope of the curve at $(x,y)$ is $-\dfrac xy=-\cot\theta$. There is, of course, a formula for the slope in terms of $dr/d\theta$, but it is not so simple.

I don't understand the problem with doing this in cartesian coordinates. We start with $x^2+y^2=2Cx$, we find the slope to be $$\frac{dy}{dx} = \frac{C-x}y = \frac{y^2-x^2}{2xy}.$$ The orthogonal trajectories will be found by solving $$\frac{dy}{dx} = \frac{2xy}{x^2-y^2}.$$ This is a homogeneous ordinary differential equation. Substitute $y=ux$, so $$x\frac{du}{dx}+ u = \frac{2u}{1-u^2},$$ and we end up with the separable differential equation $$\frac{1-u^2}{u(1+u^2)}du = \frac{dx}x.$$ This is not so bad. The left-hand side simplifies to $\dfrac 1u - \dfrac{2u}{1+u^2}$, and so we integrate and obtain $$\log u - \log (1+u^2) = \log x + c', \quad\text{i.e.,}\quad \frac u{1+u^2}=cx.$$ Not so surprisingly, this turns into the family of circles $x^2+y^2 = \frac1c y$.

EDIT: With a bit of work, one can show that $$\frac{dr}{d\theta} = r \frac{1+m\tan\theta}{m-\tan\theta}$$ where $m=dy/dx$ is the slope of the curve. One can't miss the resemblance to the addition formula for tangent here! This can be rewritten as $\displaystyle{\frac1r\frac{dr}{d\theta} = \frac1{\tan(\phi-\theta)}}$, where $\tan\phi = m = dy/dx$. Now put in $-1/m$ for $m$ and you'll get the orthogonal trajectory with $$\frac{dr}{d\theta} = -r\frac{m-\tan\theta}{1+m\tan\theta} = -r \frac1{\text{original }dr/d\theta}.$$ (Note that this is replacing $\tan\phi$ with $\tan(\phi+\pi/2) = -1/(\tan\phi)$.)

Everything is wrong here. Note that if you take a circle, say, $r=2$, then $dr/d\theta=0$. And yet the slope of the curve at $(x,y)$ is $-\dfrac xy=-\cot\theta$. There is, of course, a formula for the slope in terms of $dr/d\theta$, but it is not so simple.

I don't understand the problem with doing this in cartesian coordinates. We start with $x^2+y^2=2Cx$, we find the slope to be $$\frac{dy}{dx} = \frac{C-x}y = \frac{y^2-x^2}{2xy}.$$ The orthogonal trajectories will be found by solving $$\frac{dy}{dx} = \frac{2xy}{x^2-y^2}.$$ This is a homogeneous ordinary differential equation. Substitute $y=ux$, so $$x\frac{du}{dx}+ u = \frac{2u}{1-u^2},$$ and we end up with the separable differential equation $$\frac{1-u^2}{u(1+u^2)}du = \frac{dx}x.$$ This is not so bad. The left-hand side simplifies to $\dfrac 1u - \dfrac{2u}{1+u^2}$, and so we integrate and obtain $$\log u - \log (1+u^2) = \log x + c', \quad\text{i.e.,}\quad \frac u{1+u^2}=cx.$$ Not so surprisingly, this turns into the family of circles $x^2+y^2 = \frac1c y$.

EDIT: With a bit of work, one can show that $$\frac{dr}{d\theta} = r \frac{1+m\tan\theta}{m-\tan\theta}$$ where $m=dy/dx$ is the slope of the curve. One can't miss the resemblance to the addition formula for tangent here! This can be rewritten as $\displaystyle{\frac1r\frac{dr}{d\theta} = \frac1{\tan(\phi-\theta)}}$, where $\tan\phi = m = dy/dx$. Now put in $-1/m$ for $m$ and you'll get the orthogonal trajectory with $$\frac{dr}{d\theta} = -r\frac{m-\tan\theta}{1+m\tan\theta} = -r^2\frac1{\text{original }dr/d\theta}.$$ (Note that this is replacing $\tan\phi$ with $\tan(\phi+\pi/2) = -1/(\tan\phi)$.)

Everything is wrong here. Note that if you take a circle, say, $r=2$, then $dr/d\theta=0$. And yet the slope of the curve at $(x,y)$ is $-\dfrac xy=-\cot\theta$. There is, of course, a formula for the slope in terms of $dr/d\theta$, but it is not so simple.

I don't understand the problem with doing this in cartesian coordinates. We start with $x^2+y^2=2Cx$, we find the slope to be $$\frac{dy}{dx} = \frac{C-x}y = \frac{y^2-x^2}{2xy}.$$ The orthogonal trajectories will be found by solving $$\frac{dy}{dx} = \frac{2xy}{x^2-y^2}.$$ This is a homogeneous ordinary differential equation. Substitute $y=ux$, so $$x\frac{du}{dx}+ u = \frac{2u}{1-u^2},$$ and we end up with the separable differential equation $$\frac{1-u^2}{u(1+u^2)}du = \frac{dx}x.$$ This is not so bad. The left-hand side simplifies to $\dfrac 1u - \dfrac{2u}{1+u^2}$, and so we integrate and obtain $$\log u - \log (1+u^2) = \log x + c', \quad\text{i.e.,}\quad \frac u{1+u^2}=cx.$$ Not so surprisingly, this turns into the family of circles $x^2+y^2 = \frac1c y$.

EDIT: With a bit of work, one can show that $$\frac{dr}{d\theta} = r \frac{1+m\tan\theta}{m-\tan\theta}$$ where $m=dy/dx$ is the slope of the curve. One can't miss the resemblance to the addition formula for tangent here! This can be rewritten as $\displaystyle{\frac1r\frac{dr}{d\theta} = \frac1{\tan(\phi-\theta)}}$, where $\tan\phi = m = dy/dx$. Now put in $-1/m$ for $m$ and you'll get the orthogonal trajectory with $$\frac{dr}{d\theta} = -r\frac{m-\tan\theta}{1+m\tan\theta} = -r \frac1{\text{original }dr/d\theta}.$$

Everything is wrong here. Note that if you take a circle, say, $r=2$, then $dr/d\theta=0$. And yet the slope of the curve at $(x,y)$ is $-\dfrac xy=-\cot\theta$. There is, of course, a formula for the slope in terms of $dr/d\theta$, but it is not so simple.

I don't understand the problem with doing this in cartesian coordinates. We start with $x^2+y^2=2Cx$, we find the slope to be $$\frac{dy}{dx} = \frac{C-x}y = \frac{y^2-x^2}{2xy}.$$ The orthogonal trajectories will be found by solving $$\frac{dy}{dx} = \frac{2xy}{x^2-y^2}.$$ This is a homogeneous ordinary differential equation. Substitute $y=ux$, so $$x\frac{du}{dx}+ u = \frac{2u}{1-u^2},$$ and we end up with the separable differential equation $$\frac{1-u^2}{u(1+u^2)}du = \frac{dx}x.$$ This is not so bad. The left-hand side simplifies to $\dfrac 1u - \dfrac{2u}{1+u^2}$, and so we integrate and obtain $$\log u - \log (1+u^2) = \log x + c', \quad\text{i.e.,}\quad \frac u{1+u^2}=cx.$$ Not so surprisingly, this turns into the family of circles $x^2+y^2 = \frac1c y$.

EDIT: With a bit of work, one can show that $$\frac{dr}{d\theta} = r \frac{1+m\tan\theta}{m-\tan\theta}$$ where $m=dy/dx$ is the slope of the curve. One can't miss the resemblance to the addition formula for tangent here! This can be rewritten as $\displaystyle{\frac1r\frac{dr}{d\theta} = \frac1{\tan(\phi-\theta)}}$, where $\tan\phi = m = dy/dx$. Now put in $-1/m$ for $m$ and you'll get the orthogonal trajectory with $$\frac{dr}{d\theta} = -r\frac{m-\tan\theta}{1+m\tan\theta} = -r \frac1{\text{original }dr/d\theta}.$$ (Note that this is replacing $\tan\phi$ with $\tan(\phi+\pi/2) = -1/(\tan\phi)$.)

Everything is wrong here. Note that if you take a circle, say, $r=2$, then $dr/d\theta=0$. And yet the slope of the curve at $(x,y)$ is $-\dfrac xy=-\cot\theta$. There is, of course, a formula for the slope in terms of $dr/d\theta$, but it is not so simple.

I don't understand the problem with doing this in cartesian coordinates. We start with $x^2+y^2=2Cx$, we find the slope to be $$\frac{dy}{dx} = \frac{C-x}y = \frac{y^2-x^2}{2xy}.$$ The orthogonal trajectories will be found by solving $$\frac{dy}{dx} = \frac{2xy}{x^2-y^2}.$$ This is a homogeneous ordinary differential equation. Substitute $y=ux$, so $$x\frac{du}{dx}+ u = \frac{2u}{1-u^2},$$ and we end up with the separable differential equation $$\frac{1-u^2}{u(1+u^2)}du = \frac{dx}x.$$ This is not so bad. The left-hand side simplifies to $\dfrac 1u - \dfrac{2u}{1+u^2}$, and so we integrate and obtain $$\log u - \log (1+u^2) = \log x + c', \quad\text{i.e.,}\quad \frac u{1+u^2}=cx.$$ Not so surprisingly, this turns into the family of circles $x^2+y^2 = \frac1c y$.

Everything is wrong here. Note that if you take a circle, say, $r=2$, then $dr/d\theta=0$. And yet the slope of the curve at $(x,y)$ is $-\dfrac xy=-\cot\theta$. There is, of course, a formula for the slope in terms of $dr/d\theta$, but it is not so simple.

I don't understand the problem with doing this in cartesian coordinates. We start with $x^2+y^2=2Cx$, we find the slope to be $$\frac{dy}{dx} = \frac{C-x}y = \frac{y^2-x^2}{2xy}.$$ The orthogonal trajectories will be found by solving $$\frac{dy}{dx} = \frac{2xy}{x^2-y^2}.$$ This is a homogeneous ordinary differential equation. Substitute $y=ux$, so $$x\frac{du}{dx}+ u = \frac{2u}{1-u^2},$$ and we end up with the separable differential equation $$\frac{1-u^2}{u(1+u^2)}du = \frac{dx}x.$$ This is not so bad. The left-hand side simplifies to $\dfrac 1u - \dfrac{2u}{1+u^2}$, and so we integrate and obtain $$\log u - \log (1+u^2) = \log x + c', \quad\text{i.e.,}\quad \frac u{1+u^2}=cx.$$ Not so surprisingly, this turns into the family of circles $x^2+y^2 = \frac1c y$.

EDIT: With a bit of work, one can show that $$\frac{dr}{d\theta} = r \frac{1+m\tan\theta}{m-\tan\theta}$$ where $m=dy/dx$ is the slope of the curve. One can't miss the resemblance to the addition formula for tangent here! This can be rewritten as $\displaystyle{\frac1r\frac{dr}{d\theta} = \frac1{\tan(\phi-\theta)}}$, where $\tan\phi = m = dy/dx$. Now put in $-1/m$ for $m$ and you'll get the orthogonal trajectory with $$\frac{dr}{d\theta} = -r\frac{m-\tan\theta}{1+m\tan\theta} = -r \frac1{\text{original }dr/d\theta}.$$