I would be very grateful if someone could explain how you go out finding the form of the solution to this problem?
Have your tried expanding the solution is a Fourier series?
$\displaystyle \displaystyle \Delta T(x,y)=\left(\frac{\partial T^2}{\partial r^2} +\frac{1}{r}\frac{\partial T}{\partial r}+\frac{1}{r^2}\frac{\partial T^2}{\partial \theta^2}\right)=0$
in polar coordinates
Now use separation of variables with $\displaystyle T(r,\theta)=R(r)\Theta(\theta)$
I think we are on the same page, but just to be sure
if you separate the equations you get
$\displaystyle r^2R''+rR'-\lambda^2R=0$ and
$\displaystyle \Theta''+\lambda^2\Theta=0$
So
$\displaystyle R(r)=c_1r^\lambda+c_2r^{-\lambda}$
Since the solution must be bounded as r goes to infinity $\displaystyle c_1=0$
This gives
$\displaystyle R(r)=c_2r^{-\lambda}$
Solving the theta equation gives
$\displaystyle \Theta(\theta)=d_1\sin(\lambda \theta)+d_2\cos(\lambda \theta)$
So
$\displaystyle T_\lambda(r,\theta)=r^{-\lambda}[d_1\sin(\lambda \theta)+d_2\cos(\lambda \theta)]$
Also note that $\displaystyle \displaystyle \cos^2(\theta)=\frac{1}{2}+\frac{1}{2}\cos(2\theta )$
Use this and orthogonality to get the solution.