Suppose I have two points $A(x_A,y_A)$ and $B(x_B,y_B)$.
How can I find coefficients $a, b, c$ of the straight line general equation ?
$a x + b y + c = 0$
[Math] How to find coefficients a, b, c given two points
geometrytrigonometry
Related Solutions
An identity that might prove useful in this problem is $$ \cot\left(\frac{\theta}{2}\right)=\frac{\sin(\theta)}{1-\cos(\theta)}=\frac{1+\cos(\theta)}{\sin(\theta)}\tag{1} $$ In $\mathbb{R}^3$, one usually uses the cross product to compute the sine of the angle between two vectors. However, one can use a two-dimensional analog of the cross product to do the same thing in $\mathbb{R}^2$.
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In the diagram above, $(x,y)\perp(y,-x)$ and so the $\color{#FF0000}{\text{red angle}}$ is complementary to the $\color{#00A000}{\text{green angle}}$. Thus, $$ \begin{align} \sin(\color{#FF0000}{\text{red angle}}) &=\cos(\color{#00A000}{\text{green angle}})\\[6pt] &=\frac{(u,v)\cdot(y,-x)}{|(u,v)||(y,-x)|}\\[6pt] &=\frac{uy-vx}{|(u,v)||(x,y)|}\tag{2} \end{align} $$ $uy-vx$ is the normal component of $(u,v,0)\times(x,y,0)$; thus, it is a two dimensional analog of the cross product, and we will denote it as $(u,v)\times(x,y)=uy-vx$.
Let $u_a=\frac{Q-P}{|Q-P|}$ and $u_b=\frac{R-P}{|R-P|}$, then since $$ |A-P|=|B-P|=r\cot\left(\frac{\theta}{2}\right) $$ we get $$ \begin{align} A &=P+ru_a\cot\left(\frac{\theta}{2}\right)\\ &=P+ru_a\frac{1+u_a\cdot u_b}{|u_a\times u_b|} \end{align} $$ and $$ \begin{align} B &=P+ru_b\cot\left(\frac{\theta}{2}\right)\\ &=P+ru_b\frac{1+u_a\cdot u_b}{|u_a\times u_b|} \end{align} $$ where we take the absolute value of $u_a\times u_b$ so that the circle is in the direction of $u_a$ and $u_b$.
Given disjoint circles, and unequal radii, the locus of centers comprises two hyperbolas. Begin by intersecting the axis with the both circles. Let it intersect circle $A$ at $A_1$ and $A_2$, and circle $B$ at $B_1$ and $B_2$, as shown here, where $A_1$ and $B_1$ are between the two centers.
Let $K$ be the midpoint of $A_2B_2$, and $L$ the midpoint of $A_1B_1$. Let $P$ be the center of a circle externally tangent to both or internally tangent to both. This relation follows:
$(PA - PB)^2 = (r_a-r_b)^2$
The locus of $P$ is a hyperbola with foci $A$ and $B$. Points $K$ and $L$ both satisfy the condition for $P$, and they lie on the axis, so those are the vertices.
Now start again. Let $M$ be the midpoint of $A_2B_1$, and $N$ the midpoint of $A_1B_2$. Let $Q$ be the center of a circle externally tangent to one of the given circles and internally tangent to the other. This relation follows:
$(QA - QB)^2 = (r_a+r_b)^2$
The locus of $Q$ also is a hyperbola with foci $A$ and $B$. This time the vertices are at $M$ and $N$.
Other cases to investigate would be intersecting circles or congruent circles.
Best Answer
The answer is $$(y_A-y_B)x-(x_A-x_B)y+x_Ay_B-x_By_A=0.$$