Tangent points on circle that placed on Earth surface I need help about spherical geometry problem that I need to use it for my project.
I try to calculate $T_1 $ and $T_2$ coordinates on $B$ centered small circle on the sphere and $AT_1$ and $AT_2$ circular lines are tangent to that small circle.
Given Earth Angle Coordinates (Longitude (-180,+180) , Latitude (-90,90)) are  $A=(x_1, y_1)$  and $B=(x_2,y_2)$ . And the perimeter of circle (r) on the Earth surface is also given. 
I want to calculate $T_1 $ and $T_2$ angle coordinates.(Please suppose that Earth is perfect sphere)

My Strategy to solve the problem:


*

*I suggest that Point $A$ is my new North Pole point (0,90). 
$$B'=(0,90-\alpha)$$


Where $\alpha$ is angle between Point $A$ and point $B$.
$\alpha=|$Shortest distance between $A$ and $B$ on sphere$|/R$
$\alpha$ on unit sphere can be calculated via Spherical_law_of_cosines easily.   


*I need to calculate angle $a$



Thus
$$\sin(\alpha)=\frac{x}{R}$$
$$x=R.\sin(\alpha)$$
$$r=a.x$$
$$a=\frac{r}{x}=\frac{r}{R.\sin(\alpha)}$$


*To find relative coordinates of  coordinates $T'_1 $ and $T'_2$  are:


$$T'_1=(-a, 90-\alpha) $$ 
$$T'_2=(a, 90-\alpha)$$
I am not sure to add relative point of A and to get real coordinates because  $AB$ circle may not pass point $N$ as shown on first picture above. 
Could you please help me how to solve that problem . Thanks a lot for helps 
 A: $\DeclareMathOperator{\proj}{proj}$This can be handled easily using vector algebra. Let $A$ and $B$ be the specified points on the unit sphere, viewed as vectors in space, and assume $A \neq \pm B$ and $\cos r < A \cdot B$ ($A$ is outside the circle $C$ of radius $r$ centered at $B$). Let $U_{1}$ be the unit vector orthogonal to $B$ and tangent to the "short" great circle arc from $B$ to $A$, and let $U_{2} = B \times U_{1}$.
Theorem: If
$$
\theta = \arccos\left[\frac{(A \cdot B)\tan r}{A \cdot U_{1}}\right],
$$
then the points $T_{1}$ and $T_{2}$ are
$$
(\cos r)B + \sin r\bigl((\cos\theta) U_{1} \pm (\sin\theta) U_{2}\bigr).
$$
Computational note:


*

*$U_{1}$ is the second vector obtained from the Gram-Schmidt algorithm on the ordered basis $\{B, A\}$; that is,
$$
U_{1} = \frac{A - \proj_{B}A}{\|A - \proj_{B}A\|}.
$$


Proof of Theorem: Every point on the circle $C$ has the form
$$
T = (\cos r)B + \sin r\bigl((\cos\theta) U_{1} \pm (\sin\theta) U_{2}\bigr)
\tag{1}
$$
for some real $\theta$.
If $T$ is an arbitrary point of $C$, let $A' = A - T$ and $B' = B - T$ denote the displacements to $A$ and $B$. The projections of these vectors to the tangent plane at $T$ are
$$
A'' = A' - (A' \cdot T)T,\qquad
B'' = B' - (B' \cdot T)T,
$$
and $T$ is a point of tangency if and only if $A'' \perp B''$. Expanding the dot product, using the fact that $T$ is a unit vector $(T \cdot T = 1$) lying on the circle of radius $r$ centered at $B$ ($B \cdot T = \cos r$), gives (omitting most of the algebra)
$$
0 = A'' \cdot B''
  = A' \cdot B' - (A' \cdot T)(B' \cdot T)
  = A \cdot B - (\cos r)A \cdot T,
$$
or
$$
A \cdot T = \frac{A \cdot B}{\cos r}.
\tag{2}
$$
Substituting (1) in (2), using the fact that $A \cdot U_{2} = 0$, and solving for $\theta$ gives the value in the theorem.

