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

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  • $\begingroup$ See artofproblemsolving.com/community/c6h306012 $\endgroup$ – lab bhattacharjee Feb 13 '16 at 11:19
  • $\begingroup$ Please show me how to proceed. $\endgroup$ – Smarta Feb 13 '16 at 13:35
  • $\begingroup$ 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. $\endgroup$ – Jean Marie Feb 13 '16 at 13:59
  • $\begingroup$ 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. $\endgroup$ – Narasimham Feb 13 '16 at 14:31
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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.

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