You are right that, given the definition of the doc product, $\vec{a} \cdot \vec{b}=ab\cos\theta$, its direct application is to find the angle between the two vectors. However, such definition has many useful properties, such as the distributive one below,
$$\vec{x}\cdot (\vec{a} - \vec{b})=\vec{x}\cdot\vec{a} -\vec{x}\cdot\vec{b} $$
which in turn allows for a lot of other applications. For example, for a triangle with vertexes $\vec{a}$, $\vec{b}$ and $\vec{c}$, we have
$$\vec{c} = \vec{a} - \vec{b}$$
One could use the dot product and its distributive property to derive the cosine rule fairly effortlessly,
$$\vec{c}\cdot\vec{c} = (\vec{a}-\vec{b})\cdot (\vec{a}-\vec{b})$$
$$c^2= a^2 -2\vec{a}\cdot \vec{b} + b^2$$
$$c^2= a^2 + b^2 -2ab\cos\theta$$
Note that the proof of the Pythagorean formula is just a special case.
The derivation of the cosine rule is a convincing example for illustrating the usefulness of the dot product. There are many other useful and important applications as well in the vector space and with Cartesian coordinates.
I see two ways to consider the issue (A and B) :
A) If the coordinates of points $P_k$ are $(\cos \theta_k, \sin \theta_k)$ for $k=1,2$, consider the barycentric expression :
$$\binom{x}{y}=(1-\lambda) \binom{\cos \theta_1}{\sin \theta_1}+\lambda\binom{\cos \theta_2}{\sin \theta_2} \tag{1}$$
taking all values of $\lambda$ between $0$ and $1$, you will browse all your chord.
Said otherwise, each point of the chord will be characterized by a unique $\lambda$.
Therefore the "generic" distance to the origin will be :
$$d_{\lambda}=\sqrt{x^2+y^2}$$
with $(x,y)$ given by (1).
Remark : In order to understand formula (1), two examples,
taking $\lambda = \frac12$ you get the midpoint of your chord,
taking $\lambda = \frac34$, you are 3 times closer to $P_2$ than to $P_1$
Besides, if you eliminate $\lambda$ between the two expressions $x=..., y=...$, you will get the equation of the straight line, but I don't think that it is what you want.
B) Here is a different way.
Call $I$ the midpoint of your chord.
Let $(r_I,\theta_I)$ be the polar coordinates of $I$ (meaning in particular that distance $OI=r_I$.
If we consider now polar coordinates $(r_M,\theta_M)$ of any point $M$ on the chord, we have the simple relationship:
$$dist(O,M)=r_M=\dfrac{r_I}{\cos(\theta_I-\theta_M)}\tag{2}$$
(proof : definition of $\cos$ in right triangle $OIM$).
Best Answer
You do not have enough information. take a sheet of paper, put it on a table such that one corner $A$ is fixed, and the paper can rotate around $A$. Can you tell me what the angle is between the paper and the edge of the table? I have $A$, $L_1$ and $L_2$, and those are perpendicular, but the angle can take any value since I can rotate the paper.