Please refer to the following diagram:
Let the center of the circle be oriented at the origin of an $xy$-coordinate system. Let the line $\overline{AC}$ lie along the line given by:
$$y=\cot(\alpha)x+k$$
Substituting for $y$ into the equation of the circle, we obtain:
$$x^2+\left(\cot(\alpha)x+k \right)^2=r^2$$
Expanding, we find:
$$\left(1+\cot^2(\alpha) \right)x^2+2k\cot(\alpha)x+k^2-r^2=0$$
Using the Pythagorean identity $$\csc^2(\theta)=\cot^2(\theta)+1$$ we may write:
$$\csc^2(\alpha)x^2+2k\cot(\alpha)x+k^2-r^2=0$$
Multiplying through by $$\sin^2(\alpha)$$ we get:
$$x^2+2k\sin(\alpha)\cos(\alpha)x+ \sin^2(\alpha)\left(k^2-r^2 \right)=0$$
Applying the quadratic formula, the roots of this quadratic are then given by:
$$x=\frac{-2k\sin(\alpha)\cos(\alpha)\pm \sqrt{\left(2k\sin(\alpha)\cos(\alpha) \right)^2-4(1) \left(\sin^2(\alpha)\left(k^2-r^2 \right) \right)}}{2\cdot1}$$
Simplifying, we obtain:
$$x=\frac{-2k\sin(\alpha)\cos(\alpha) \pm2\sin(\alpha)\sqrt{\left(k\cos(\alpha) \right)^2- \left(k^2-r^2 \right)}}{2}$$
$$x=\sin(\alpha)\left(-k\cos(\alpha)\pm\sqrt{k^2\cos^2(\alpha)-k^2+r^2} \right)$$
$$x=\sin(\alpha)\left(-k\cos(\alpha)\pm\sqrt{r^2-k^2\sin^2(\alpha)} \right)$$
The smaller of these roots is the $x$-coordinate of point $C$ and the larger is the $x$-coordinate of point $B$, hence:
$$b=r-\sin(\alpha)\left(-k\cos(\alpha)+\sqrt{r^2-k^2\sin^2(\alpha)} \right)=r+\sin(\alpha)\left(k\cos(\alpha)-\sqrt{r^2-k^2\sin^2(\alpha)} \right)$$
$$c=r-\sin(\alpha)\left(-k\cos(\alpha)-\sqrt{r^2-k^2\sin^2(\alpha)} \right)=r+\sin(\alpha)\left(k\cos(\alpha)+\sqrt{r^2-k^2\sin^2(\alpha)} \right)$$
Adding these two equations, we get:
$$b+c=2r+2k\sin(\alpha)\cos(\alpha)$$
Solving this for $k$, there results:
$$k=\frac{b+c-2r}{2\sin(\alpha)\cos(\alpha)}$$
Substituting for $k$ into the expression equal to $b$, we find:
$$b=r+\sin(\alpha)\left(\left(\frac{b+c-2r}{2\sin(\alpha)\cos(\alpha)} \right)\cos(\alpha)-\sqrt{r^2-\left(\frac{b+c-2r}{2\sin(\alpha)\cos(\alpha)} \right)^2\sin^2(\alpha)} \right)$$
Simplifying, we find:
$$b=\frac{b+c}{2}-\sin(\alpha)\sqrt{r^2-\left(\frac{b+c-2r}{2\cos(\alpha)} \right)^2}$$
$$\sqrt{r^2-\left(\frac{b+c-2r}{2\cos(\alpha)} \right)^2}=\frac{c-b}{2\sin(\alpha)}$$
Squaring:
$$r^2-\left(\frac{b+c-2r}{2\cos(\alpha)} \right)^2=\left(\frac{c-b}{2\sin(\alpha)} \right)^2$$
Multiply through by $$4\cos^2(\alpha)$$:
$$4\cos^2(\alpha)r^2-\left(b+c-2r \right)^2=\left((c-b)\cot(\alpha) \right)^2$$
$$4\cos^2(\alpha)r^2-(b+c)^2+4r(b+c)-4r^2=\left((c-b)\cot(\alpha) \right)^2$$
$$4\sin^2(\alpha)r^2-4(b+c)r+(b+c)^2+\left((c-b)\cot(\alpha) \right)^2=0$$
$$4\sin^2(\alpha)r^2-4(b+c)r+b^2+2bc+c^2+\cot^2(\alpha)\left(c^2-2bc+b^2 \right)=0$$
$$4\sin^2(\alpha)r^2-4(b+c)r+\csc^2(\alpha)\left(b^2+c^2 \right)+2bc\left(1-\cot^2(\alpha) \right)=0$$
Applying the quadratic formula, and taking the smaller of the two roots, we obtain:
$$r=\frac{4(b+c)-\sqrt{16(b+c)^2-16\sin^2(\alpha)\left(\csc^2(\alpha)\left(b^2+c^2 \right)+2bc\left(1-\cot^2(\alpha) \right) \right)}}{2\left(4\sin^2(\alpha) \right)}$$
$$r=\frac{b+c-\sqrt{(b+c)^2-\left(\left(b^2+c^2 \right)-2bc\cos(2\alpha) \right)}}{2\sin^2(\alpha)}$$
$$r=\frac{b+c-\sqrt{2bc\left(1+\cos(2\alpha) \right)}}{2\sin^2(\alpha)}$$
$$r=\frac{b+c-2\cos(\alpha)\sqrt{bc}}{2\sin^2(\alpha)}$$
