Determine the diameter length (mm) of each circle shown using the given information. Geometry, trig and algebra only. 
I have considered ratios of triangles as well constructing tangent and secants to compare sides of triangles. Neither has worked out for me. Also have considered moving small circle to center but that has not been advantageous.
 A: Call $R$ and $r$ the radius of the big and small circles repectively. Since $2r+90=2R$ we get $R=r+45$
Now notice $DB=DG=r=R-45$. Also notice $DG=\sqrt{GC^2+DC^2}$. 
notice $GC=R-50$ and $DC=R-r=45$ so we get $GC^2+DC^2=R^2-100R+2500+45^2$.
So $\sqrt{R^2-100R+2500+2025}=DG=DB=R-45\implies$
$ R^2-100R+2500+2025=R^2-90R+2025$. 
Notice the last part is a quadratic in $R$, so from here we can obtain $R$ which happens to be $250$.
And from here $r=R-45=205$
A: If the big circle's radius is $\;R\;$ and the small circle's is $\;r\;$ , and if $\;D=(0,Y_D)\;$ , then you can check that $\;Y_D=r-R<0\;$
Since the point $\;E:=(0, Y_E)\;$ is common to both circles, then
$$\begin{cases}&Y_E^2=R^2\\{}\\&(Y_E-Y_D)^2=r^2\end{cases}$$
and since clearly $\;Y_D\,,\,\,Y_E<0\;,\;\;R,r>0\;,\;\;R>|Y_D|$ ,  we get
$$Y_E=-R\;,\;\;(-R-YD)^2=(R+Y_D)^2=r^2$$
Now,
$$90=Y_A-Y_B=R-(Y_D+r)=R-r-Y_D=-2Y_D\implies Y_D=-45$$
Try now to take it from here.
A: Consider a circle around $C$, through $H$ which should be between $F$ and $G$, so that $DE = CH$ -- a circle the same size as the small one, centered at the center of the big one.  This gives $FH = 45mm$ (because the difference in radiuses is clearly $45mm$), so $GH = 5mm$.  This gives us something we can use: $GC = r-5$, $CD = 45$, and $GD = r$, and those form a right triangle.  A quick run through the pythagorean theorem and some solving finishes the job.
