Question

Find the locus of a point P if the tangents drawn from P to circle $x^2+y^2=a^2$. so that the tangents are perpendicular to each other.

Answer 1

Let $\overrightarrow{p}$ be the position if point $P$. For a given $\hat{x}$ it intersect the circle if ${|\overrightarrow{p}+\mu \lambda |}^2=a^2$ for some value for $u\epsilon R$

${\mu }^2+2\overrightarrow{p}.\hat{x}\mu +P^2=a^2\Rightarrow$ { $\mu =-\overrightarrow{P}.\hat{x}$

{ ${(\hat{p}.\hat{x})}^2=P^2$

Therefore two solutions ${\hat{x}}_1$ for $\hat{x}$. So the intersection occurs at two points $\overrightarrow{p}-\overrightarrow{P}.\hat{x}-\hat{x}$ and $\overrightarrow{P}-\overrightarrow{p}.{\hat{x}}_{+}.{\hat{x}}_{+}$. Locus definite yields

$O=[\overrightarrow{p}-(\overrightarrow{p}-\overrightarrow{p}.\hat{x}-\hat{x})][\overrightarrow{p}-(\overrightarrow{p}-\overrightarrow{p}.{\hat{x}}_{+}{\hat{x}}_{+})]=(\overrightarrow{p}.{\hat{x}}_{-})(\overrightarrow{p}.{\hat{x}}_{+}){\hat{x}}_{-}.{\hat{x}}_{+}$

$p=\left|\overrightarrow{p}\right|=\sqrt{2}a$

$x^2+y^2={2a}^2$.