>>13845314It's not that hard.
The equation of the top right circle is
(x-5)^2 + (y-2)^2 = 1
So the tangents of the circle are the lines
cos(t) x + sin(t) y = 5 cos(t) + 2 sin(t) + 1
with a parameter t.
To find the one who goes through the origin we just have to find t so that (x,y) = (0,0) is on the line so
5 cos(t) + 2 sin(t) + 1 = 0
This has two solution so we pick the one where t is between 0 and pi namely
t = arctan((2 - 10 sqrt(7))/(5 + 4 sqrt(7))) + pi
Now we just calculate the intersection points between
(x-3)^2 + (y-2)^2 = 1
cos(t) x + sin(t) y = 5 cos(t) + 2 sin(t) + 1
which are
x1 = -2*((206*sqrt(7))/841 + 1084/841 + (4*sqrt(5594 + 2683*sqrt(7)))/841)*(sqrt(7) - 5)/3
y1 = (206*sqrt(7))/841 + 1084/841 + (4*sqrt(5594 + 2683*sqrt(7)))/841
x2 = -2*((206*sqrt(7))/841 + 1084/841 - (4*sqrt(5594 + 2683*sqrt(7)))/841)*(sqrt(7) - 5)/3
y2 =(206*sqrt(7))/841 + 1084/841 - (4*sqrt(5594 + 2683*sqrt(7)))/841
and use Pythagoras to get there distance which is
sqrt((x1 - x2)^2 + (y1 - y2)^2)
= sqrt(107648 + 1022656*sqrt(7))/841
which is approximately
1.9944135994157613262257425687963320956066918545365
