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$$(x-x_1)(x-x_2) + (y-y_1)(y-y_2) + \lambda| \begin{bmatrix} x&y&1\\x_1 & y_1 & 1 \\ x_2 &y_2 &1 \end{bmatrix}| = 0$$

The above equation is supposed to represent family of all circles passing through two points, but I can't understand why it should be. I understand so far that the first two terms i.e: the $(x-x_1)(x-x_2) + (y-y_1)(y-y_2)$ part represents the diametric form of circle and the determinant is the equation of line passing through $(x_1,y_1)$ and $(x_2,y_2)$ but I don't understand why we needed to add line to get family of all circles passing through the point. Isn't the first two terms enough?

Clemens Bartholdy
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2 Answers2

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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.

Sam Freedman
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  • I didn't the angle sin(<ACB) thing , could you explain a bit more – Clemens Bartholdy Feb 06 '21 at 20:25
  • I think I missed a factor of 1/2. Consider the parallelogram spanned by $CA$ and $CB$; its area is base * height. Base is $|CB|$, and the height is $|CA| \sin(\theta)$, where $\theta$ is the included angle. – Sam Freedman Feb 06 '21 at 20:30
  • After reading it again, the tetrahedron is a 3-d object if I am not mistaken but we are talking about planar stuff, so it's kinda weird to think like this for me – Clemens Bartholdy Feb 06 '21 at 20:31
  • Agreed that it’s weird! You can think of the plane as $z=1$ in 3D space. The vectors from the origin to this plane form the edges of the tetrahedron. – Sam Freedman Feb 06 '21 at 20:32
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We can use $$ \cot\theta=\frac{\left(1+m_{1}m_{2}\right)}{m_{2}-m_{1}}=k $$ which simplifies to $$ \frac{\left(x-x_{1}\right)\left(x-x_{2}\right)+\left(y-y_{2}\right)\left(y-y_{1}\right)}{\left(y-y_{1}\right)\left(x-x_{2}\right)-\left(y-y_{2}\right)\left(x-x_{1}\right)}=k $$ See(3-point form of circle)

simplifying and replacing $k$ with $\lambda$ gives us the required result.

PP's here
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