Find elevation of a tangent in a circle

Question

I am trying to understand how I can calculate an elevation (i.e. the distance) of a tangent line given an arc and radius.

For example :

Given that I know $d$ and $s$, how do I get the value for $?$ ?

If the value $s$ will never be greater than $\frac{\pi}{2}$, then when $s = \frac{\pi}{2}$, the value for $? = d$, right? So can this be solved as a linear function?

(Note: while I have sufficient understanding of maths, this is not my best subject.)

Edit

Sorry about the picture. I made this as a 2 minutes drawing in Gimp; $s$ is the distance of an arc. (I should've placed that variable within the circle, instead...)

Answer 1

Let $L$ denote the length of the hypotenuse. Then

$$ L=d\sec\left(\dfrac{s}{d}\right)$$

So the excess length is

$$ ?=L-d=d\left( \sec\left(\dfrac{s}{d}\right)-1\right)$$

Note that the angle is here measured in radians.

Answer 2

If $\theta = s/d $ is angle upto red line starting and $T = d \tan \theta $ is the tangent length of segment. (I use x instead of ?). By circle tangent/secant property

$$ x (2 d +x) = T^2 = (d \tan \theta )^2$$

Solve the quadratic equation for $x$ discarding one root that is not relevant.

$$ x = \sqrt { d^2+ {T^2} } -d $$

$$ \frac{x}{d} = \sqrt { 1+ \tan{^2}\theta } -1 = \sec \theta -1 $$

Actually this is straight trig!!

Answer 3

The distance of the intersection point from the center of the circle is $d \sec \theta = \frac{d}{\cos \theta}$, where $\theta$ is the angle subtended by the arc. Now, $\theta = \frac{s}{d}$, so the desired altitude is

$$ d \sec (s/d) -d = d \left[\sec (s/d)-1\right] $$

Note: Using $d$ for the radius instead of the diameter is a bit confusing! :-)