What's the power of point $P $ in relation to the center the circumference $O2$?

For reference:

Given the externally tangent circles of centers $O_1$ and $O_2$, by $O_1$ the lines $L_1$ and $L_2$ ($L_1 \perp L_2$) are drawn so that $L_1$ is tangent to the circle of center $O_2$ at $T$ and $L_2$ intercepts the circumference of center $O_1$ in $N$ and $M$ and L2 is tangent to the circle of center O2 . Calculate Pot($M/O_2$), if $MN = 10$ and Pot($M/O_2$) < Pot($N/O_2$). (Answer:$50$)

My progress: I'm not sure my drawing is correct.

$\triangle O_1O_2G:$

Th.Pit: $R^2+R^2=(R+5)^2 \implies R = 5+5\sqrt2$

Th. Median: $(10+5\sqrt2)^2+(5+5\sqrt2)^2 = 2(O_2M)^2+\frac{(10+5\sqrt2)^2}{2} \implies (O_2M)^2 =100+50\sqrt2\\ Pot(M/O_2)=MG^2 = (R-O_1)^2 =(5+5\sqrt2-5)^2 = 50 $

Where am I going wrong?

Given that $MN=10$, the radius of $\odot O_1$ is equal to $5$.

From Pythagoras' theorem to $\triangle O_1O_2G$,

$(R+5)^2=2R^2\implies R^2=25+10R$

Also, $MG=O_1G-OM=R-5$.

Power of point $M$ is, $$(MO_2)^2-(O_2G)^2=MG^2=(R-5)^2=R^2-10R+25=25+\require{cancel}\cancel{10R}-\cancel{10R}+25=50.$$