Equation to the circle.

Question

How to show that the equation to the circle of which the points $(x_1,y_1)$ and $(x_2,y_2)$ are the ends of a cord of a segment containing an angle $\theta$ is, $$(x-x_1)(x-x_2)+(y-y_1)(y-y_2) ± \cot(\theta)[(x-x_1)(y-y_2)-(x-x_2)(y-y_1)]=0$$

Answer 1

I have had a look at the reference you have given, @lab bhattacharjee. It is exactly the same problem wording. The solutions given there are interesting. I provide another one which, in my opinion, provides a more direct path.

Let $M_k(x_k,y_k)$. Let $O$ be the center of the circle. Let us assume that angle $(\overrightarrow{OM_1},\overrightarrow{OM_2})=2 \theta$.

Point $M$ belongs to the circle if and only if $(\overrightarrow{MM_1},\overrightarrow{MM_2})=\theta$ (half angle property). This constraint can be interpretated in the following way:

$\dfrac{|\overrightarrow{MM_1}.\overrightarrow{MM_2}|}{\| \overrightarrow{MM_1}\times \overrightarrow{MM_2}\|}$ $=\dfrac{\|\overrightarrow{MM_1}\|\|\overrightarrow{MM_2}\|(\pm\cos{\theta })}{\|\overrightarrow{MM_1}\|\|\overrightarrow{MM_2}\|\sin{\theta }}=\pm\dfrac{\cos{\theta}}{\sin{\theta}}=\pm\cot{\theta}$.

It suffice now to turn to the analytical expressions of the dot product and the cross product to obtain the given equation.