This is easy to formulate as a semidefinite programming problem.  

First, let $X=xx^{T}$.  The semidefiniteness constraint becomes

$A-\lambda X \succeq 0$

Next, use a standard technique to handle the absolute value in the objective by replacing it with an auxiliary variable and two linear inequality constraints.  The problem becomes 
 
$\min_{\lambda,t} t $
 
subject to 
 
$t \geq \lambda-\lambda_{0} $

$t \geq \lambda_{0}-\lambda $
 
$A-\lambda X \succeq 0$
 
If $t$ is greater than or equal to $\lambda-\lambda_{0}$ and $t$ is greater than or equal to $\lambda_{0}-\lambda$, then $t$ is clearly greater than or equalt to $| \lambda-\lambda_{0} |$.  Since $t$ is being minimized and there are no other constraints on $t$, it will end up equal to $| \lambda-\lambda_{0}|$. 
 
This isn't quite in standard SDP format.  The two constraints involving $t$ can be brought into semidefinite form by making 

$t - \lambda + \lambda_{0} $
 
and

$t - \lambda_{0} + \lambda $
 
diagonal elements of the matrix that is constrained to be positive semidefinite.  This insures that $t-\lambda+\lambda_{0} \geq 0$ and $t-\lambda_{0}+\lambda \geq 0$. 
 
Let 

$
F_{0}=\left[ 
\begin{array}{ccc}
A & 0 & 0 \\\
0 & \lambda_{0} & 0 \\\
0 & 0 & -\lambda_{0}
\end{array}
\right]
$ 

$
F_{1}=\left[
\begin{array}{ccc}
-X & 0 & 0 \\\
0  & -1 & 0 \\\
0  & 0  & 1  
\end{array}
\right]
$

$F_{2}=\left[
\begin{array}{ccc}
0 & 0 & 0 \\\
0 & 1 & 0 \\\
0 & 0 & 1 
\end{array}
\right]
$

Now, the problem can be written in standard form as

$\min_{\lambda,t} t $
 
subject to 

$F_{0}+\lambda F_{1}+tF_{2} \succeq 0$