Background
The flow field around a half-infinite body is defined by a potential flow consisting of a parallel flow with velocity $u = U_{\infty}$ and a source of magnitude Q at position $x = a$, $y = b$
Parallel flow situation: we have $u = U_{\infty}$ and $v = 0$,
$$\phi_{paralell}(x,y) = u_{\infty}x + v_{\infty}y \Rightarrow \phi(x,y) = ux$$
Source flow situation: we have $u = \frac{Q}{2\pi}\frac{x}{x^2+y^2}$ and $v = \frac{Q}{2\pi}\frac{y}{x^2+y^2}$,
$$\phi_{source}(x,y) =\frac{Q}{2\pi} \ln (r) =\frac{Q}{2\pi} \ln (\sqrt{x^2+y^2})$$
$$ \implies \phi_{paralell} + \phi_{source} = \phi_{total}$$
With a source with magnitude ”Q” located at $x = a$, $y = b$. The superimposed velocity potential reads: $$ u_{\infty}(x-a) + \frac{Q}{2\pi} \ln \Big(\sqrt{(x-a)^2+(y-b)^2}\Big) $$ We can then calculate the velocity field $\bar{u} = (u,v)^T$ from the velocity potential $\phi$
$$u = \frac{\partial \phi}{\partial x}=u_{\infty} +\frac{Q}{2\pi} \frac{(x-a)}{(x-a)^2 + (y-b)^2} \quad \textbf{(1)}$$
and $$v = \frac{\partial \phi}{\partial y} = \frac{Q}{2\pi} \frac{(y-b)}{(x-a)^2 + (y-b)^2}$$
The problem
Calculate the position of all stagnation points $(u = v = 0)$. For the following questions assume that the source is placed at the origin, $a = b = 0$. Use cylindrical coordinates, $x = r \cos(\theta)$ and y = $r \sin(\theta)$ for your calculations.
The answer At the stagnation point ($SP$) the velocity is zero, $u = v = 0$. Evaluating this condition for the velocity field found above results for
the x-direction in: \begin{align} u &= 0 \\ \iff 0 &= u_{\infty} + \frac{Q}{2\pi } \frac{1}{x_{SP} - a} \quad \textbf{(2)}\\ \iff x_{SP} &= -\frac{Q}{2\pi u_{\infty}} + a \quad \textbf{(3)} \end{align}
the y-direction in: $$v = 0 \iff y_{SP} = b$$
Here is what I don't understand
how can we go from $\textbf{(1)}$ to $\textbf{(2)}$ and $\textbf{(3)}$? What are the mathematical and physical explanations behind
