In LC circuit, the EM waves specifically originate from where? In LC circuit, we know that there will be charge oscillations, so emw will be produced. I just want to know emw will originate from where? Will it originate from within the capacitor?
Or will it originate from the wires? (Like EM waves due to AC source.)
 A: It will originate from everywhere: the wires, the capacitor and all the elements carrying the current or the displacement current. Indeed, thee emission of the EM waves is described by the Maxwell equations where the sources are the time-dependent currents and fields.
What is confusing in LC circuit is that it is not suitable for the description of the electromagnetic wave generation, since it is a simlplification, where all the electromagnetic properties are collapsed into point-like elements, such as capacitance, inductance, resistance, etc. On the other hand, waves are extended in space.
An important corollary is that spatial extent of the circuit limits the length of the EM waves that it can generate. This is why one often uses antennas, with a good rule of the thumb that the length of the antenna is about half of the wave length (in more technical terms,  the antenna is needed for impedance matching between the circuit and the space where the EM wave propagates, assuring that maximum of energy is transferred from the circuit to the propagating wave.)
A: EM waves are generated in a circuit having a physical inductor and capacitor but in an ideal "$LC$" circuit charges move and oscillate without EM waves being generated. The reason for no waves is the assumption that the circuit consists of infinitesimally small elements and connections, in practice, having linear extent negligible compared to $\lambda_0=c/f_0$, where $f_0$ is the frequency of oscillation. A real $LC$ circuit of finite extent will generate and radiate EM waves, the larger its size the more efficiently. For example, a magnetic dipole antenna is nothing but a coil of wire. If you resonate its inductance with a capacitor the circuit will radiate at the resonant frequency but the amount of radiation will be proportional to the ratio $(D/\lambda_0)^4$ where $D$ is the coil's diameter, as $D\to 0$ this ratio becomes quickly $0$. The same goes with a capacitor, an electric dipole radiator, whose radiation is proportional to $(d/\lambda_0)^2$ where $d$ is the distance between the capacitor plates, again as $d\to 0$ this too quickly becomes negligible but not as fast as the magnetic dipole radiator. Of course, for an ideal $LC$ both $D=0$ and $d=0$.
You can see the effect of radiation by measuring the mutual coupling of coils or capacitors. Two coils couple through their mutual inductance as in a transformer, or if you place two capacitors side-by-side you get their capacitive couplings. As you increase their distance the coupling goes down. In real life, this coupling could be very undesirable because it can lead to unexpected and erratic circuit behavior the most dramatic being radiation from the amplified output end of an amplifier can be picked up at (i.e., couple to) the input of the amplifier via inductive and/or capacitive elements and drive the amplifier into oscillation and potentially total failure.
Note too that in a real life $LC$ even the connecting wires (even those traces on a PC board) can and do radiate and RF engineers spend a lot of time to minimize that spurious radiation to interfere with the circuit's desired behavior.
A: The radiation comes from the wires. If you solve for the field of the wires you will find a near field proportional to the current, which falls off as r$^{-2}$, and a radiation field proportional to the time derivative of the current, which falls off as  r$^{-1}$. See Jackson, Griffiths or Barut.
