Given the family of curves $x=r\cos(\theta)$ and $y=r\sin(\theta)$ with $\theta \in [0,2\pi)$, I wish to find another family of curves orthogonal to it. One method is as follows: $$x^2+y^2=r^2 \implies 2x+2y\frac{dy}{dx}=0 \implies \frac{dy}{dx}=-\frac{x}{y}.$$ Inverting the slope to find curves orthogonal to it, we have $$\frac{dy}{dx} = \frac{y}{x} \implies y=cx,$$ as we expected. However, if we do not know the trick of writing $x^2+y^2=r^2$ and instead brute force via substitution $$x=r\cos\left(\arcsin\left(\frac{y}{r}\right)\right) \implies \frac{dx}{dy}=-\frac{y}{r\sqrt{1-\frac{y^2}{r^2}}}.$$ Inverting the differential gives $$\frac{dx}{dy}=\frac{r\sqrt{1-\frac{y^2}{r^2}}}{y}\implies x=r\sqrt{1-\frac{y^2}{r^2}}-r\tanh^{-1}\left(\sqrt{1-\frac{y^2}{r^2}}\right)+c,$$ which does not recover the expected result. The first term is just $x$, so what is left is a constant $c$ equal to a function of $x$, which is not true. Can anyone see why writing $x$ in terms of $y$ directly won't lead to an orthogonal curve?
Hence I considered a simple case. Suppose $x=f(r), y=g(r)$ are some more complicated parametric curves. What we typically do is to perform implicit differentiation on the function $h(x,y,r)=0$ and then invert the slope to get a differential equation for the orthogonal curves. It seems like the brute force approach above is following this procedure, but somehow cannot lead to the correct answer.