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

• – lab bhattacharjee Feb 13 '16 at 11:19
• Please show me how to proceed. – Smarta Feb 13 '16 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. – Jean Marie Feb 13 '16 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. – Narasimham Feb 13 '16 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:
$\dfrac{|\overrightarrow{MM_1}.\overrightarrow{MM_2}|}{\| \overrightarrow{MM_1}\times \overrightarrow{MM_2}\|}$ $=\dfrac{\|\overrightarrow{MM_1}\|\|\overrightarrow{MM_2}\|(\pm\cos{\theta })}{\|\overrightarrow{MM_1}\|\|\overrightarrow{MM_2}\|\sin{\theta }}=\pm\dfrac{\cos{\theta}}{\sin{\theta}}=\pm\cot{\theta}$.