# Equation to the circle.

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, $$(x-x_1)(x-x_2)+(y-y_1)(y-y_2) ± \cot(\theta)[(x-x_1)(y-y_2)-(x-x_2)(y-y_1)]=0$$

• Commented Feb 13, 2016 at 11:19
• Please show me how to proceed. Commented Feb 13, 2016 at 13:35
• You ask how to proceed. I ask you the same thing as @Lovsovs. If you are completely blocked, I can say how I have found the demonstration: I have spotted a dot product on the left hand side, and, once I looked at the right hand side, I asked myself "It looks like a determinant of the same vectors I have on the left", I can interpret a determinant as the area of the generated parallelogram, but after a moment, I realized that I had better to do by expressing it at the norm of the cross product which does the same job. Commented Feb 13, 2016 at 13:59
• Another way of stating the circle property that angle $\theta$ subtended by triangle's legs at circumference of circle is constant that can be found from dot product of the vectors.Or it may be also product of cutting line segments. Commented Feb 13, 2016 at 14:31

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