If I know coordinates of point $A$, coordinates of circle center $B$ and r as the radius of the circle, is it possible to calculate lines that are passing through point $A$ that are also tangent to the circle?
A is the green point, B is the center of the red circle and I am trying to find out the blue lines.
Denote coordinates of point $A$ with $(x_A, y_A)$ and center B with $(x_B, y_B)$. Suppose that tangent AT touches circle at point $T$ with coordinates $(x_T,y_T)$.
You have:
$$AB=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}\tag{1}$$
$$AT=\sqrt{AB^2-r^2}\tag{2}$$
Denote the angle between line AB and $x$-axis with $\alpha$ and the angle $\angle TAB$ with $\beta$. You can calculate the value of $\alpha$ from:
$$\alpha=\arctan\frac{y_B-y_A}{x_B-x_A}\tag{3}$$
...and the angle $\beta$ from the right triangle $\triangle ATB$:
$$\beta=\arcsin\frac{r}{AB}\tag{4}$$
Note that the solution exists only if (2) makes sense ($AB\ge r$). And if you are using computers to do the calculation, you should always use function like $atan2$ to compute $\alpha$ from the right quadrant.
Coordinates of point T can be obtained from the following expressions:
$$x_T=x_A+AT \space \cos(\alpha \pm\beta)$$
$$y_T=y_A+ AT \space \sin(\alpha \pm\beta)$$
In general, you have two different solutions for $AB>r$ and only one, trivial, for $AB=r$ ($T\equiv A$).
Equations of tangents are:
$$y-y_A=(x-x_A)\tan(\alpha\pm\beta)$$