How can I find the curves in the problem? It's the first exercise in this book (It has a drawing).

Let $ \rho = f(\theta) $ be the equation of a plane curve in polar coordinates.  Through the pole $O$ draw a line perpendicular to the radius vector $OM$, and let $T$ and $N$ be the points where this line cuts the tangent and the normal.  Find expressions for the distances $OT$, $ON$, $MN$, and $MT$ in terms of $f(\theta)$ and $f'(\theta)$.
Find the curves for which each of these distances, in turn, is constant.

I tried to find the curve with $OT$ being constant.
$$\begin{aligned}
M (\rho\cos \theta, \rho\sin\theta)\\
T (k\sin\theta, k\cos\theta)\\
\frac{y_T-y_M}{x_T-x_M} = \frac{dy_M}{dx_M} \\
(k\cos\theta-\rho\sin\theta)(\rho'\cos\theta-\rho\sin\theta)=\\(k\sin\theta-\rho\cos\theta)(\rho'\sin\theta+\rho\cos\theta)
\end{aligned}$$
What should I do next?
 A: We'll solve just the case for which $OT$ is constant. The missing cases you can easily complete.
Let $\vec {r}$ the position vector of point M, and $\vec{i}$ and $\vec{j}$ the versors of axes $x$ and $y$ respectively. Let the set of vectors $\{\vec{u},\vec{v}\}$ an orthonormal basis and $\rho =f(\theta)$.
See the figure below:

So
$$\vec {r} = f(\theta) \vec {u} \quad (I)$$
$$\vec {u}= \cos(\theta)\vec {i} +\sin(\theta) \vec {j}$$
$$\vec {v}= \dfrac{d\vec{u}}{d\theta} = -\sin(\theta)\vec {i} +\cos(\theta) \vec {j}$$
Let $\vec {t}$ a tangent vector at point M. The vector $\vec {t}$ can be defined as:
$$\vec {t} = \dfrac{d\vec{r}}{d\theta}.$$
So
$$\vec {t} = f'(\theta) \vec {u} + f(\theta) \vec{v}.$$
Let's calculate $OT$ now.
We know that
$$\overrightarrow{OT}=\vec {r} + \overrightarrow{MT}.\quad (1)$$
But
$$\overrightarrow{MT}= \lambda \vec{t} \quad (2)$$
and
$$\overrightarrow{OT}= \mu \vec{v}. \quad(3)$$
So using equations $(I)$, $(1)$, $(2)$ and $(3)$ we get:
$$f(\theta) \vec{u} + \lambda[f'(\theta) \vec{u} + f(\theta) \vec{v}]= \mu \vec{v} \Rightarrow$$
$$(f(\theta)+\lambda f'(\theta)) \vec{u} + \lambda f(\theta) \vec{v}= \mu \vec{v}. \quad (4) $$
It follows that
$$\lambda = - \frac{f(\theta)}{f'(\theta)} \quad (5)$$
and
$$\mu = - \frac{f^2(\theta)}{f'(\theta)}. \quad (6)$$
Therefore
$$OT= \frac{f^2(\theta)}{f'(\theta)}. \quad (7)$$
As $\triangle NMT$ is a right triangle we can easily calculate $ON$, $MN$, and $MT$.
For example:
We know that
$$ON \cdot OT = OM^2, \quad (8) $$
then using $(7)$ and $(8)$ we get:
$$ON = f'(\theta). \quad (9)$$
If we want a curve for which $OT$ is constant ($k$) then
$$\frac{f^2(\theta)}{f'(\theta)}=k \Rightarrow$$
$$\Rightarrow k \dfrac{df}{d\theta}=f^2(\theta) \Rightarrow$$
$$\Rightarrow k \frac{d\rho}{\rho^2}= d\theta \Rightarrow$$
$$\Rightarrow -k \rho^{-1}=\theta + C$$
and we are done.
