What is the Radiation Resistance after using matching network on a dipole antenna?

For example, using a λ/2 center fed dipole. Zin=73Ω+j42.5Ω where Rr=73Ω. Now if I want to get rid of the reactance, by a parallel cap across the input of the dipole.

So we parallel a cap with Y_c=+j0.005956 to cancel out the reactance so Yin=0.01023=1/Rin. This gives Rin=97.742

So Rin =97.742Ω, not 73Ω. We know current peak at the input, so what is the Radiation Resistance? If Rr is 73, then what is 97.742 and where is the extra resistance goes?

Thanks

Input current is less but input voltage is more than the current and voltage across the "radiation resistance". So, the resistance looks bigger at the input. The power in is equal to the power dissipated in the radiation resistance, though.

Thanks for the reply.
We know the current peaks at the input point. So what is the Rr? Or you mean the peak current is no longer at the feed point after impedance matching?

The antenna is like a resistor in series with an inductor. The resistance of the resistor is 73 ohms. When you add a capacitor in parallel with this series circuit, the resistance of the overall circuit is 97 ohms. The radiation resistance of the antenna is still 73 ohms, though. The effect of the capacitor and the inductor is like a transformer.... It transforms the73 ohms to 97 at the input terminals. The current distribution in the wire is not changed. The model of the antenna is not changed. You are not feeding the signal to across the 73 ohm resistor. You are feeding the signal at the junction of the capacitor and inductor. The current here is different than the current flowing into the 73 ohm resistor, because of the capacitor. You can use a circuit simulator to see how this works, you don't need a e.m. Simulator. Try it with spice, for example.