Question 792578
On an x-y coordinate plane, a circle of radius R has the equation ({{{x^2+y^2=R^2}}}
Let's say you start at point A(0,R).
If you leave the circle on the tangent at point A(0,R) and travel a distance b you get to point B(b,R).
The radius of the circle going through A is the y-axis, the line {{{x=0}}},
and the tangent at A is the perpendicular line {{{y=30.85}}}.
If you turn {{{90^o}}} towards the circle, you will travel along the line {{{x=b}}} and will get to the circle at point C.
{{{drawing(300,300,-0.99,0.99,-0.99,0.99,
grid(0),circle(0,0,0.8),
triangle(0,0.8,0.4,0.8,0.4,0.6928),
rectangle(0.4,0.8,0.37,0.77),
locate(0.02,0.79,A),locate(0.41,0.85,B),locate(0.36,0.68,C)
)}}}
Point C has {{{x=b}}}
We can find the y-coordinate from the equation of the circle.
{{{b^2+y^2=R^2}}} --> {{{b^2=R^2-b^2}}} --> {{{b=sqrt(R^2-b^2)}}}
The distance between points B and C is the difference between their y-coordinates,
{{{R-sqrt(R^2-b^2)}}}
Of course you need to make {{{b<=R}}}.
Otherwise, when you turn {{{90^o}}} at point B(b,R) you are too far to the right and pass by the circle without ever finding a point C.
 
In your case, {{{R=30.85m }}} is very large compared to {{{b=1m}}}.
The triangle ABC for that case looks so ridiculously thin that I did not help explain the problem.
The formula above calculates the distance between points B and C as
{{{30.85-sqrt(30.85^2-1^2)}}}= approximately {{{30.85-30.8338=0.0162}}}
When R is so much larger than b, you can calculate a good approximation with the formula
{{{b^2/2R}}}, which in this case would give you {{{1/2/30.85=1/61.7}}}= approximately {{{0.0162}}}