If we consider the velocity potential $\textbf{q} = k^2 \frac{x \textbf{j} - y \textbf{i}}{x^2 + y^2} , k =$ constant.
The streamlines are given by the circles $x^2 + y^2 =$ constant, the circles.
Now the streamlines cut the equipotential surfaces orthogonally.
We also found the equipotential surfaces $\phi(x,y,z) = c$ given by $q = -\nabla \phi$.
That is $\phi(x,y)= k^2 tan^{-1}(\frac{x}{y}) = c$ That is they represent the planes $x = cy$.
And also it is intuitively and geometrically clear that the streamlines cut the equipotential surface orthogonally.
But my main question is what happens at the origin, like because we have the equipotential surfaces as planes passing through the origin ? , at origin the equipotential surfaces are not even defined?
This is the velocity field induced by a line vortex and there is a singularity at the origin in the plane. Because of this singularity, the potential is, in fact, not defined at the line $\{(0,0,z): -\infty < z < \infty\}$ and the equipotential surfaces converge here.
The flow is irrotational everwhere except at $(x,y) = (0,0) $, with $\nabla \times \mathbf{q} = 0$, and where there exists a well-defined differentiable potential $\phi$ such that $\mathbf{q} = - \nabla \phi$.
It can help clarify by switching to cylindrical coordinates: $\,x = r \cos \theta, \,\, y = r \sin \theta$. The unit basis vectors are $\mathbf{e}_r = \cos \theta \,\,\mathbf{i} - \sin \theta \,\,\mathbf{j}$ and $\mathbf{e}_\theta = -\sin \theta \,\,\mathbf{i} + \cos \theta \,\,\mathbf{j}$ and the velocity field is
$$u_r = 0, \,\,\,u_\theta = \frac{k^2}{r}, \,\,\, u_z = 0.$$
Note that $u_\theta$ is infinite at $r = 0.$
The potential is given by
$$-\nabla \phi = -\left(\frac{\partial \phi}{\partial r}\,\mathbf{e}_r + \frac{1}{r} \frac{\partial \phi}{\partial \theta}\,\mathbf{e}_\theta + \frac{\partial \phi}{\partial z}\right) = \frac{k^2}{r} \mathbf{e}_\theta,$$
with solution $\phi = - k^2 \theta$ for $0 \leqslant \theta < 2\pi$. We restrict the domain of $\theta$ to $[0, 2\pi)$ so that the potential is not multi-valued for $(x,y) \in \mathbb{R}^2 \setminus (0,0)$ and $\phi$ is undefined for $(x,y) = (0,0)$.