If  $(x-a)\cos \theta +y\sin \theta =(x-a)\cos \varphi +y\sin \varphi =a$  and  $\tan (\theta /2)-\tan (\varphi /2)=2b,$  then 

Let  $\tan (\theta /2)=\alpha and\tan (\varphi /2)=\beta ,$
So that  $\alpha -\beta =2b$
Also  $\cos \theta =\frac{1-{\tan }^2(\theta /2)}{1+{\tan }^2(\theta /2)}=\frac{1-{\alpha }^2}{1+{\alpha }^2}$
and  $\sin \theta =\frac{2\tan (\theta /2)}{1+{\tan }^2(\theta /2)}=\frac{2\alpha }{1+{\alpha }^2}$
similarly  $\cos \varphi =\frac{1-{\beta }^2}{1+{\beta }^2}$ and  $\sin \varphi =\frac{2\beta }{1+{\beta }^2}$
Therefore, we have from the given relations
 $$\begin{gathered} (x-a)\frac{1-{\alpha }^2}{1+{\alpha }^2}+y\left(\frac{2\alpha }{1+{\alpha }^2}\right)=a \\ \Rightarrow x{a}^2-2y\alpha +2a-x=0 \end{gathered}$$
Similarly  $x{\beta }^2-2y\beta +2a-x=0$
We see that  $\alpha$  and  $\beta$  are the roots of the equations
 $x{z}^2-2yz+2a-x=0,$
So that  $\alpha +\beta =2y/xand\alpha \beta =(2a-x)/x$
Now, from  ${(\alpha +\beta )}^2={(\alpha -\beta )}^2+4\alpha \beta$ ,
We get  ${\left(\frac{2y}{x}\right)}^2={\left(2b\right)}^2+\frac{4(2a-x)}{x}$
 $\Rightarrow y^2=2ax-(1-b^2)x^2$
Also, from $\alpha +\beta =2y/x$  and  $\alpha -\beta =2b,$
We get  $\alpha =y/x+b$   and  $\beta =y/x-b$
$\Rightarrow \tan \frac{\theta }{2}=\frac{1}{x}(y+bx)$  and  $\tan \frac{\varphi }{2}=\frac{1}{x}(y-bx)$