Solution:

Method of image for a circle in a free stream near a wall. check for the boundary condition of no flux on the surface of the circle.

First, we obtain the $x$ and $y$ components of the velocity vector.

$$V_{y} = \frac{\partial \phi}{\partial y} = Ux\left(-\frac{2a^... ...^2)^2}-\frac{2a^{2}(y+b)}{(x^{2}+(y+b)^{2})^{2}}\right) \notag$$

and

$$V_{x} = \frac{\partial\phi}{\partial x} = U\left( {1 + \frac{... ...y-b)^2)^2}-\frac{2a^{2}x}{(x^{2}+(y+b)^{2})^{2}}\right) \notag$$

Second, we make the change of variables

$$\begin{align}x = & r\cos\theta, \notag \\ y = & b+r\sin\theta. \notag \end{align}$$

We substitute the expressions for $x$ and $y$ in terms of $r$ and $\theta$ in the equations for $v_{x}$ and $v_{y}$. Then, we set $r = a$, and we define $p = \cos\theta, q = \sin\theta$ and $t = b/a$. Now we write

$$\frac{v_{x}}{U} = \left(1+\frac{1}{p^{2}+q^{2}}+\frac{1}{p^{2... ...2}+q^{2})^{2}}-\frac{2p}{(p^{2}+(q+2t)^{2})^{2}}\right) \notag$$

and

$$\frac{v_{y}}{U} = p\left(-\frac{2q}{(p^{2}+q^{2})^{2}}-\frac{2(q+2t)}{(p^{2}+(q+2t)^{2})^{2}}\right). \notag$$

Third, evaluate the radial velocity $V_{r}$ at the circles surface. The radial velocity $V_{r}$ can be written in terms of the velocities $V_{x}$ and $V_{y}$ according to the equation

$$\begin{align}V_{r} = & V_{x}(r,\theta)\frac{\partial x}{\partial r}+V_{y}(r,\the... ... \\ = & V_{x}(r,\theta)\cos\theta+V_{y}(r,\theta)\sin\theta \notag \end{align}$$

With this equation and the equations for $V_{x}$ and $V_{y}$ at the surface of the circle, we have

$$\begin{split} V_{r}\vert _{r=a} & = p\left(1+\frac{1}{p^{2}+q... ...ac{2(q+2t)}{(p^{2}+(q+2t)^{2})^{2}}\right), \end{split} \notag$$

which simplifies to

$$V_{r}\vert _{r=a} = \frac{p(4t^{2}-1)}{[p^{2}+(q+2t)^{2}]^{2}} \notag$$