On finding the complex number satisfying the given conditions Question:- Find all the complex numbers $z$ for which $\arg\left(\dfrac{3z-6-3i}{2z-8-6i}\right)=\dfrac{\pi}{4}$ and $|z-3+i|=3$

My solution:-
$\begin{equation}
\arg\left(\dfrac{3z-6-3i}{2z-8-6i}\right)=\dfrac{\pi}{4} \implies \arg\left(\dfrac{z-2-i}{z-4-3i}\right)=\dfrac{\pi}{4}
\end{equation}\tag{1}$
Now, what this suggests is that $z$ lies on a segment with the end points of the segment being $(2,1)$ and $(4,3)$. Now to find the equation of the circle we find the center and radius of the circle which comes out to be $(4,1)$ and $2$, respectively. So, the equation of the circle comes out to be 
$$(x-4)^2+(y-1)^2=4$$
Now all we need to do is find the point of intersection of the circles $|z-4-i|=2$ and $|z-3+i|=3$ to find the desired $z$, which comes out to be 
$$\boxed {z=\left(4 \mp \dfrac{4}{\sqrt{5}}\right) + i\left(1 \pm \dfrac{2}{\sqrt{5}} \right)}$$.
Now just to check which one is the solution I substituted both these $z$ back into $(1)$ but I found out that neither satisfy the condition as the argument comes out to be $\dfrac{\pm\sqrt{5}+5}{2}$, instead of 1, so according to my solution neither are the answer.

My problem with the question:-
But as it turns out as per the book I am solving, the answers are both the complex numbers $z$ that I found out. Also, I think only one of these is the solution because on some rough plotting I found out that the circle $|z-3+i|=3$ intersects the segment $|z-4-i|=2$ with end points as mentioned in the solution only once at $z=\left(4+\dfrac{4}{\sqrt{5}}\right) + i\left(1-\dfrac{2}{\sqrt{5}} \right)$. 
So, is it me that is wrong or the book or maybe both. If there is anything wrong in my solution please point it out.

From the following where $\circ=\pi/4$, 
$\qquad\qquad\qquad $
the circle $$(x-4)^2+(y-1)^2=4$$
you've found is correct, but note that we have only the part below the line $y=x-1$ passing through $(2,1)$ and $(4,3)$. 
This is because the angle (as measured in counter-clock wise direction) from $\vec{zz_1}$ to $\vec{zz_2}$ is $\pi/4$ where $z_1=4+3i,z_2=2+i$.  
$\qquad\qquad\qquad$
Thus, solving
$$(x-4)^2+(y-1)^2=4$$
$$y\lt x-1$$
$$(x-3)^2+(y+1)^2=3^2$$
gives only one solution $$x=4+\frac{4}{\sqrt 5},\quad y=1-\frac{2}{\sqrt 5}$$
(and this satisfies $(1)$.)
