What is the area $\triangle DEF$ ?
I solved it using analityc geometry. I want to see if there is way to solve it using plane geometry. I did it: $x^2+y^2=400$
$(x+10)^2+y^2=100$
I found the radius fo the red's circle $\left(x-\frac{20}{3}\right)^2+\left(y-\frac{80}{9}\right)^2=\left(\frac{80}{9}\right)^2$
So the tangent line ax+y+c=0 common to both circle is given by the below system :
$|c-10a| = 10\sqrt{a^2+1}$
$\frac{20}{3}a+\frac{80}{9}+c| = \frac{80}{9}\sqrt{a^2+1}$
And so on ...
Thank you for your atention.
may be a mixture method will be quick way.
you already have red one center $O_2 (\dfrac{20}{3},\dfrac{80}{9}),r_2=\dfrac{80}{9}$, connect $O_1O_2$, let it's middle point is $P$,$EF$middle is $Q$,then $DPQ$ will be co-line,
(edit 2)
first check $PQ=\dfrac{O_1O_2}{2}$, put $PG\perp SO_1, G $ fall on $SO_1$ think $\triangle MPQ$ and $\triangle SPO_1$, they are similar.
you can know every thing of $\triangle SPO_1$,you know $MQ=SQ=FQ=\sqrt{r_1r_2}$
now you can write down $DP$ since you know the angle of $MQ\perp O_1O_2$ and you can find out $D$ quickly. now find $|DP|$, it is easy.