In the figure determine $'x'$ , knowing that $PM=MQ=4$ and $O$ and $O'$ are centers.(S:$x=4$)
I try:
$\triangle OQH \sim \triangle OAP \implies \dfrac{HO}{OP} = \dfrac{HQ}{AP} = \dfrac{8+OP}{AO}$
$\triangle (OQH-PA): 8.AO.NH = OP.NQ.AH$
$\triangle PQN \sim \triangle HAN: \dfrac{HN}{PN}=\dfrac{AN}{QN}=\dfrac{AH}{PQ}$
$\triangle HAN \sim \triangle QHO: \dfrac{HN}{OH}=\dfrac{AN}{8+OP}=\dfrac{AH}{QH}$
Draw the triangle $\triangle OPD$ and $\triangle OLP$
By simple angle chasing you can find out the blue angles are equal.
Now look at $\triangle QLO$ and $/triangle GLO$ they are similar.
With this information, you can get the following relationship
$\frac{r-4}{OL}=\frac{r+4}{OL}$
$(OL)^2=(r+4).(r-4)$
Look at $\triangle QLN$ as $\angle QLN=90^o$
$r^2=x^2+(OL)^2$ as $(OL)^2=(r+4).(r-4)$
Then finally u get,
$r^2=x^2+(r+4).(r-4)$
$r^2=x^2+r^2-4^2$
AT THE END
$\boxed{x=4}$
Thank you @peta arantes for giving us a great problem to solve!