Question

Prove that the equation to the circle of which the points $(x_1,y_1)$ and $(x_2,y_2)$ are the ends of a chord of a segment containing an angle $\theta$ is $(x-x_1)(x-x_2)+(y-y_1)(y-y_2)\pm \cot \theta \left[\right(x-x_1\left)\right(y-y_2)-(x-x_2\left)\right(y-y_1\left)\right]=0.$

Answer 1

Let the point $A,B,$ and $C$ be $(h,k),(x_1,y_1)$ and $(x_2,y_2)$ respectively

Let slope of $AB$ be $m_1$ and slope of $AC$ be $m_2$

$m_1=\frac{k-y_1}{h-x_1}{m}_2=\frac{k-y_2}{h-x_2}\tan \theta =\left|\frac{m_1-m_1}{1+m_1{m}_2}\right|\tan \theta =\pm \frac{\frac{k-y_1}{h-x_1}-\frac{k-y_2}{h-x_2}}{1+\frac{k-y_1}{h-x_1}\times \frac{k-y_2}{h-x_2}}\frac{1}{\cot \theta }=\pm \frac{(k-y_1)(h-x_2)-(h-x_1)(k-y_2)}{(h-x_1)(h-x_2)+(k-y_1)(k-y_2)}(h-x_1)(h-x_2)+(k-y_1)(k-y_2)=\pm \cot \theta \left(\right(k-y_1\left)\right(h-x_2)-(h-x_1\left)\right(k-y_2\left)\right)(h-x_1)(h-x_2)+(k-y_1)(k-y_2)\pm \cot \theta \left(\right(k-y_1\left)\right(h-x_2)-(h-x_1\left)\right(k-y_2\left)\right)=0$

Generalising the equation

$(x-x_1)(x-x_2)+(y-y_1)(y-y_2)\pm \cot \theta \left\{\right(y-y_1\left)\right(x-x_2)-(x-x_1\left)\right(y-y_2\left)\right\}=0$

Hence proved.