Given circle A (center point and radius $R$ is given), and a line segment (2 points are given: $P_{0}$ and $P_{1}$).
$P_{0}$ is on the circle.
There is another circle with given radius $r$, and it is tangential on the circle and the line segment.
What is the circle's location? (there are 2 possible locations)
Let $P_1=(x_1,y_1),\space P_0=(x_0,\sqrt{R^2-x_0^2})$; the center should be in the circle $x^2+y^2=(R-r)^2$
By geometry of the figure, the center $C$ can be found as follows:
$(1)$ Ligne $P_1P_0:ax+by=c$ where $a,b,c$ are known. So there are two distant parallel of $r$ from the $P_1P_0$ whose equations are $$ax+by=c\pm r\sqrt{a^2+b^2}$$ $(2)$ The center is given by the system $$\begin{cases}ax+by=c\pm r\sqrt{a^2+b^2}\\x^2+y^2=(R-r)^2\end{cases}$$ EXAMPLE.- $(R,r)=(5,1),\space P_0=(3,4), P_1=(-2,1)$ and line $P_0P_1:\space 3x-5y=-11$
Parallel lignes to a distance $r:3x-5y=-11\pm\sqrt{34}$
Solutions of the two systems $(x,y)=(0.889,3.9)$ and $(x,y)=(2.889,2.767)$ So we get the two circles $$(x-0.889)^2+(y-3.9)^2=1 \text { and } (x-2.889)^2+(y-2.767)^2=1$$ which appear in the attached figure.