Family of circles passing through two given points

Solution 1:

Given two points $A = (x_1, y_1)$ and $B = (x_2, y_2)$, we want to find a circle for which $AB$ is a chord, not necessarily a diameter. Choose $C = (x, y)$ on this circle. Because the inscribed angle $\angle ACB$ is independent of the choice of $C$ (it depends only on the subtended arc), we get an equation $$\frac{CA \cdot CB}{|CA||CB|} = k$$ for some constant $k$.

(When $AB$ is a diameter, $\angle ACB$ is a right angle. Then $k= 0$ and you get the diametric equation for a circle.)

In any case, we get $$(x_1 - x)(x_2 - x) + (y_1 - y)(y_2 - y) = k|CA||CB|.$$

Here's where my argument gets somewhat inelegant, but I think it still works:

Note that $\frac12|CA||CB|\sin(\angle ACB)$ is the area of $\Delta ABC$. Moreover, the determinant in the equation is the volume of a tetrahedron with base $\Delta ABC$ and height $1$, i.e. $1/3$ of the area of $\Delta ABC$. Since $\sin(\angle ACB)$ is a constant depending only on the circle, you can now solve for the $\lambda$ to get the third parameter.