Set the center of the circle to be $\left(0,0\right)$. Then the set of points on the circle are those such that $x^2+y^2=r^2$ and the set of points on the line segment are solutions to $y=mx+b$ from some $m$ and $b$ (these are easily found given the two endpoints of the line segment). Substituting, we have
\begin{equation}
x^2+(mx+b)^2=r^2\Longleftrightarrow (m^2+1)x^2+2mbx+(b^2-r^2)=0
\end{equation}
Then this is just a quadratic you can solve with the quadratic formula giving you $x$, and you can get $y$ by $y=mx+b$. However, this gives you two solutions; you must pick the right one (find which is between the two endpoint). Fortunately finding the angle after this is pretty easy: it's just simple trigonometry. If $(x,y)$ is the solution, $\tan^{-1}\left(\frac{y}{x}\right)$ gives you the angles measure counterclockwise from the $x$-axis. You asked for the angle from the $y$-axis, which is then $\frac{\pi}{2}-\tan^{-1}\left(\frac{y}{x}\right).$
A worked example
I'll work out the example you have given. We have radius $r=\frac{3}{2}$, center $\left(3,4\right)$ and a line segment from $(3,3)$ to $(6,7)$. First we'll move the center to $(0,0)$. We do this by subtracting $(3,4)$ from each point. This gives us the center (as desired) of $(0,0)$, and a line segment from $(0,-1)$ to $(3,3)$. Then the line between the two points is $y=\frac{4}{3}x-1$. The equation for the circle is $x^2+y^2=\frac{9}{4}$. Substituting the equation for the line in, we have
\begin{equation}
x^2+\left(\frac{4}{3}x-1\right)^2=\frac{25}{9}x^2-\frac{8}{3}x+1=\frac{9}{4}
\end{equation}
which gives the quadratic
\begin{equation}
\frac{25}{9}x^2-\frac{8}{3}x-\frac{5}{4}=100x^2-96x-45=0
\end{equation}
Then the quadratic equation gives $x=\frac{96\pm \sqrt{96^2+4\cdot 100\cdot 45}}{200}=-0.345,1.305$. We're going to want the solution with positive $x$, so $x=1.305$. Then $y=\frac{4}{3}(1.305)-1=0.740$. To find the desired angles, we merely take $90-\tan^{-1}\left(\frac{0.740}{1.305}\right)=90-29.56=60.44$. I'm not sure what the discrepancy between my answer and the one you provided is; it could be rounding errors, but it is more likely I just made a calculation error somewhere (please do point it out if you see it).
I know it seems like a lot, but the technique is relatively straight forward, the calculation is just some hard work. I hope this answer has shed some light on the process for you.
The centers ($R$ and $S$ below) of the target circles lie on the parabola whose focus is the tangent point ($T$) and directrix is the "other" tangent line (through some point $U$).
As the figure shows, one readily determines that the corresponding radii ($r$ and $s$) satisfy
$$r(1+\cos\theta) = d = s(1-\cos\theta) \quad\to\quad \{r,s\}=\frac{d}{1\pm\cos\theta}\tag{1}$$
where $d$ is the focus-directrix distance (ie, the distance from the tangent point to the "other" tangent line) and $\theta$ is the angle between direction vectors of the two lines (equivalently, between normal vectors).
Now, if $t$ and $u$ are the unit normal vectors to the tangent lines through $T$ and $U$, then we can write
$$d = |(U-T)\cdot u| \qquad \cos\theta = t\cdot u \tag2$$
so that
$$\{R,S\} = T \pm_1 \frac{|(U-T)\cdot u|}{1\pm t\cdot u} \;t \tag3$$
where $\pm_1$ is a hedge on my part in case of a sign error due to reversing a vector somewhere. Barring an error, you "should" be able to leverage the signed distance $d$ to write
$$\{R,S\} = T \pm \frac{(U-T)\cdot u}{1\pm t\cdot u}\;t \tag4$$
where the two $\pm$s match.
(Sanity check: Changing the direction of $t$ effectively reverses each $\pm$, so that's consistent. Likewise for $u$. So the only question is whether I should've written $T-U$ instead of $U-T$, but I believe my GeoGebra sketch bears me out on that one.)
Note that, if the lines are parallel, then $t\cdot u = \pm 1$ (with the sign depending upon how the normals are chosen), so that one of the centers $R$, $S$ lies halfway between the lines, and the other shoots off to infinity, as expected. $\square$
Best Answer
\begin{align} \tan\alpha&=\frac{|FQ|}{|PQ|} =\frac{|IB_t|}{|PB_t|} \tag{1}\label{1} . \end{align}
Let $|PQ|=x$. Then
\begin{align} |PB_t|&= \sqrt{|PI|^2-|IB_t|^2} =\sqrt{(x+d+r)^2-r^2} . \end{align}
From \eqref{1} \begin{align} \frac dx&= \frac{r}{\sqrt{(x+d+r)^2-r^2}} ,\\ \frac{d^2}{x^2}&= \frac{r^2}{(x+d+r)^2-r^2} =\frac{r^2}{x^2+2x(d+r)+d(d+2r)} , \end{align}
\begin{align} (d^2-r^2)\,x^2+2\,d^2\,(d+r)\,x+d^3\,(d+2r)&=0 , \end{align}
\begin{align} x&= \frac{d^2(d+r+r\sqrt{2(1+r/d)})}{r^2-d^2} ,\\ \tan\alpha&= \frac dx= \frac{r^2-d^2}{d(d+r+r\sqrt{2(1+r/d)})} \\ &= \frac{1-s^2}{s^2+s+\sqrt{2\,s\,(s+1)}} . \end{align} Now, show that
\begin{align} \frac{1-s^2}{s^2+s+\sqrt{2\,s\,(s+1)}} - \frac{1+s-s\,\sqrt{2s(s+1)}}{s+(1+s)\sqrt{2s(s+1)}} \equiv0 . \end{align}