Question

Show that the locus of the point the tangents from which to the circle $x^2+y^2=a^2$ include a constant angle $\alpha$ is $(x^2+y^2-2a^2{)}^2{\tan }^2\alpha =4a^2(x^2+y^2-a^2)$.

Answer 1

$\angle PTC=\alpha /2$ where $T$ is $(h,k)$ and where
$\tan \frac{\alpha }{2}=\frac{CP}{PT}=\frac{a}{\sqrt{(h^2+k^2-a^2)}}.....\left(1\right)$
$\tan \alpha =\frac{2\tan \left(\alpha /2\right)}{1-{\tan }^2\left(\alpha /2\right)}=\frac{2a\sqrt{(h^2+k^2-a^2)}}{h^2+k^2-a^2-a^2}$
Squaring both sides, we get
$(h^2+k^2-2a^2{)}^2{\tan }^2\alpha =4a^2(h^2+k^2-a^2)$.
Hence the required locus of the point is
$(x^2+y^2-2a^2{)}^2{\tan }^2\alpha =4a^2(x^2+y^2-a^2)$
Particular case : If the tangents be at right angles than $\alpha ={90}^{\circ }\therefore \alpha /2={45}^{\circ }$
Hence from $\left(1\right)$ we get $\sqrt{(h^2+k^2-a^2)}=a$ or $h^2+k^2=2a^2$
$\therefore$ Required locus is $x^2+y^2=2a^2$.