The reason the $x$ and $y$ components lie on circles is because the expectation values for the components are zero (here you should check the math in the book you're learning from):
$$\langle L_x\rangle=\langle L_y\rangle=0$$
When one talks about expectation values then one should always consider an experiment. In your case an experiment would be that we first prepare your state as shown in the picture, where $L_z$ can take on any value $m\hbar$. Next we measure the $L_x$ and $L_y$ components. If we repeat this - preparation, measuring - a gazillion times then we get as an average value for the $x$ and $y$ component the value $0$ for both.
This means that after you prepared your state in a state of definite $L_z$ and then measure the $x$ and $y$ components that these can point in any direction if you measure them. This is why the $x$ and $y$ components lie on circles: then for each possibility that one component points the one way there is an equal probability that this component points the other way. So for each $L_z$ both $L_x$ and $L_y$ lie on a circle. If you had only discrete points then the expectation values would not be zero.
When you prepare your state such that $L_x$ has definite values then you get the same picture as above, only with $x$ instead of $z$. But then my argument is the same, because if you want the $x$ component to be precise then
$$\langle L_z\rangle=\langle L_y\rangle=0$$
and both the $z$ and $y$ components lie on circles. So, everytime you know that one component has the form $m\hbar$, where $-l\leq m \leq l$, you know that the others are not of this form.