Circle , Prove that the locus of the point of intersection of the lines drawn through the points $(a,0)$ and $(-a,0)$

Question

Prove that the locus of the point of intersection of the lines drawn through the points $(a,0)$ & $(-a,0)$ which include a constant angle $\theta$ is the Circle $x^2 +y^2 -a^2 -2ay\cot \theta =0$ OR $x^2 +y^2 -a^2 +2ay\cot \theta =0$

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Answer 1

Note that $\angle MKB = \theta$.

$MK$ and $KB$ can then be expressed in terms of $a$ and $\theta$.

Use the center-radius form to write the equation of the circle. It should be $$x^2 + (y – a \cot \theta)^2 = (a \csc \theta)^2.$$

Part of the required result follows after simplification of the above.

The other part of the required is the lower half of the circle using $\pi – \theta$ as the constant angle.