I am able to find common tangent lines of two circles in 2D space ( point is x,y ).
but the problem is can I use the equation to be used in spherical surfaces ( point is lat, lon )? , in the sense that x is longitude and y is latitude but still using great circle distance as the distance equation between two points
Consider the problem spatially. Let $\vec a$ and $\vec b$ be the centers of the circles on the surface of the sphere. (They are unit vectors in $\mathbb R^3$). The points $\vec p$ that lie on the first circle satisfy:$$ \lVert \vec p \rVert = 1 \\ \vec p \cdot \vec a = \cos A $$
$A$ (and $B$) are the angles that describe the size of the given circles.
A great circle is specified by its axis, $\vec v$ (the normal of the plane the great circle lies on). The tangency condition is a constraint on the angle between vector $\vec v$ and vectors $\vec a$ and $\vec b$: $$ \vec v \cdot \vec a = \pm \sin A\\ \vec v \cdot \vec b = \pm \sin B\\ \lVert \vec v \rVert^2 = 1 $$
Substitute:$$ \vec v=x \vec a+ y \vec b + z (\vec a \times \vec b)\\ x \vec a \cdot \vec a + y \vec a \cdot \vec b = \pm \sin A\\ x \vec a \cdot \vec b + y \vec b \cdot \vec b = \pm \sin B\\ \lVert x \vec a+ y \vec b + z (\vec a \times \vec b) \rVert = 1 $$ Simplifying: $$ x + y \vec a \cdot \vec b = \pm \sin A\\ x \vec a \cdot \vec b + y = \pm \sin B\\ x^2 + y^2 + z^2 \lVert \vec a \times \vec b\rVert^2 + 2xy \vec a \cdot \vec b = 1 $$ The $\pm$ are independent -- they specify which sides of the circles your great circle is tangent to. $$ S_a=\pm \sin A\\ S_b=\pm \sin B\\ x = \frac{S_a - S_b \vec a \cdot \vec b}{1-(\vec a \cdot \vec b)^2}\\ y = \frac{S_b - S_a \vec a \cdot \vec b}{1-(\vec a \cdot \vec b)^2} $$ You can then solve for $z$, and thus, find $\vec v$