Find relation between $a,b,c,d$ such that the graph of $f(x) = {{ax+b}\over{cx+d}}$ is a shifted graph of $g(x) = {1\over x}$ 
Given a function $f(x) = {{ax+b}\over{cx+d}}$ find the relation between $a,b,c,d$ such that the graph of $f(x)$ is a shifted graph of $g(x) = {1\over x}$

I'm trying to find a way to attack that problem. What i've done so far is as follows:
We know that $g(x)$ is an odd function and is therefore symmetric with respect to $(x=0, y=0)$. That fact means that its shifted version will also be symmetric with respect to some point $(x_0, y_0)$. Based on that we may assume that:
$$
{{ax+b}\over{cx+d}} = {1 \over {x + x_0} } + y_0
$$
By making algebraic transformations i got an equation:
$$
ax^2 + (ax_0 + b)x+bx_0=cy_0x^2+(c+dy_0+cx_0y_0)x+(d+dx_0y_0)
$$
From the equation we may get a system of equations for the coefficients and try to find the connection between them:
\begin{cases}
  a=cy_0\\      
  ax_0+b=c+dy_0+cx_0y_0\\
  bx_0=d+dx_0y_0
\end{cases}
We may get rid of $x_0$ and $y_0$ by expressing them: $y_0 ={a\over c}$, $x_0 = {{cd+a}\over{c(b-d)}}$.
So eventually I got:
$$
b\cdot{{cd+a}\over{c(b-d)}} = d\cdot\left(1+{a\over c}\cdot {{cd+a}\over{c(b-d)}} \right)
$$
And that doesn't match with the answer in the book which is $c^2 = bc-ad$.
Seems that my reasoning above is wrong. What am i missing?
 A: 
By making algebraic transformations i got an equation:

It's not completely clear to me what steps you took there; but here's an alternative.
Adding the step in blue:

$$
{{ax+b}\over{cx+d}} = {1 \over {x + x_0} } + y_0 \color{blue}{= \frac{y_0x+1+x_0y_0}{x+x_0}}
$$

Now divide numerator and denominator of the initial fraction by $c$ $(*)$ and compare the corresponding coefficients in:
$${{\color{red}{\tfrac{a}{c}}x+\color{purple}{\tfrac{b}{c}}}\over{x+\color{green}{\tfrac{d}{c}}}} = \frac{\color{red}{y_0}x+\color{purple}{1+x_0y_0}}{x+\color{green}{x_0}}$$
to get:
$$\left\{\begin{array}{l}
\tfrac{a}{c} = y_0 \\
\tfrac{b}{c} = 1+x_0y_0 \\
\tfrac{d}{c} = x_0
\end{array}\right.$$
and substitution of the first and last equation into the middle one yields:
$$\frac{b}{c} = 1+\frac{a}{c}\frac{d}{c} \iff bc=c^2+ad$$

$(*)$ Note that assuming $c \ne 0$ is no problem since for $c=0$, the expression $\tfrac{ax+b}{cx+d}=\tfrac{ax+b}{d}$ could never be a shifted version of $\tfrac{1}{x}$.
A: If $c\neq 0$ and $bc-ad\neq 0$, then $\frac{ax+b}{cx+d}= \frac{a}{c}+\frac{b-ad/c}{c(x+d/c)} = \frac{a}{c}+\frac{1}{\frac{c^2}{bc-ad}\cdot(x+d/c)}$. 
So the condition seems to be $\frac{c^2}{bc-ad}=1$, or equivalently $c^2=bc-ad$. 
Clearly $c\neq 0$ is necessary. 
If $bc-ad= 0$, then $\frac{ax+b}{cx+d}= \frac{a}{c}$ by the above rearrangement. 
To sum up: $c^2=bc-ad\neq 0$ is the equivalent condition. 
