The equation of a generic circle is $x^2+y^2+ax+by+c=0$. In order for it to pass through point $(1,0)$ the condition $1+a+c=0\Rightarrow c=-a-1$ (obtained by substituting the coordinates of the point in the circle equation) needs to be satisfied.
The two conditions of tangency are respected when the system of equation composed by the circle equation and that of the considered line gives a solution. Hence, the two systems below: \begin{equation} \begin{cases} x^2+y^2+ax+by+c=0\\ x=8\end{cases} \Rightarrow 64+y^2+8a+by+c=0 \end{equation} \begin{equation} \begin{cases} x^2+y^2+ax+by+c=0\\ x=0\end{cases} \Rightarrow y^2+by+c=0 \end{equation} Since each of these 3 conditions must be satisfied at the same time, we write another system of equation: \begin{equation} \begin{cases} 64+y^2+8a+by+c=0\\y^2+by+c=0\\ c=-a-1\end{cases} \end{equation} By solving it, I get: \begin{equation} \begin{cases} 64+y^2+8a+by-a-1=0\\y^2+by-a-1=0\\ c=-a-1\end{cases} \Leftrightarrow \begin{cases}y^2+7a+by+63=0\\by=-y^2+a+1\\ c=-a-1\end{cases} \end{equation} \begin{equation} \Leftrightarrow \begin{cases} 8a+64=0\\by=-y^2+a+1\\ c=-a-1\end{cases} \Leftrightarrow \begin{cases} a=-8\\y^2+by+7=0\\ c=7\end{cases} \end{equation} At this point I'm stuck since from the second order algebraic equation $y^2+by+7=0$ I should infer the value of $b$ thus solving the problem but I still have the unknown $y$.
I need some help, please!
Since it is tangent to $x=0$ and $x=8$ its center must be of the form $(4,y_0)$ and its radius must be $4.$ So, its equation is
$$(x-4)^2+(y-y_0)^2=16.$$ Since the point $(1,0)$ belongs to the circle we have that $$9+y_0^2=16.$$ That is $y_0=\pm\sqrt 7.$ Thus there are two circles satisfying the given conditions $$(x-4)^2+(y-\sqrt 7)^2=16$$ and $$(x-4)^2+(y+\sqrt 7)^2=16.$$