Finding 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? 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.
I tried solving this and then I got to this condition here, after I applied the formulua for finding the angle between the tangents
Formula is Angle btw tangents: $$\cos\theta = \frac{1 - \tan^2(\theta/2)}{ 1 + \tan^2(\theta/2)} $$
So, I got to this equation of locus after solving using that formula...
$$a^2 * \cos^2(\theta/2) = x_1^2 + y_1^2$$
But I am having trouble trying to figure out how to show that the tangents are perpendicular :C
so, I tried applying the trigonometric here, and then I got this answer
$$x_1^2 + y_1^2 - a^2 * \cos^2 (\theta/2)$$
But in my solutions book it's different, it's $x_1^2 + y_1^2 - 2a^2$ 
 A: HINT.
Tangent lines are perpendicular and equal between them. They are also perpendicular to the radii passing throuh tangency points. So tangent lines and radii form a square.
A: Hint: find the hidden square in the picture below.

A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Leftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[2]{\,\mathrm{Li}_{#1}\left(\,{#2}\,\right)}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
Lets $\ds{\vec{p}}$ the position of point $P$. For a given $\ds{\hat{n}}$, it intersects the circle if
$\ds{\verts{\vec{p} + \mu\hat{n}}^{2} = a^{2}}$ for some value of $\ds{\mu \in \mathbb{R}}$:
\begin{align}
\mu^{2} + 2\vec{p}\cdot\hat{n}\,\mu + p^{2} = a^{2} & \imp\
\left\lbrace\begin{array}{rcl}
\ds{\mu} & \ds{=} &
\ds{-\vec{p}\cdot\hat{n}}
\\[1mm]
\ds{\pars{\vec{p}\cdot\hat{n}}^{2} + a^{2}} & \ds{=} & \ds{p^{2}}
\end{array}\right.\tag{1}
\end{align}
There are $\ds{\ul{two}}$ solutions $\ds{\hat{n}_{\pm}}$ for $\ds{\hat{n}}$. So, the intersection ocurrs at two points:
$\ds{\vec{p} - \vec{p}\cdot\hat{n}_{-}\,\hat{n}_{-}}$ and
$\ds{\vec{p} - \vec{p}\cdot\hat{n}_{+}\,\hat{n}_{+}}$. Then, the locus definition yields
$$
0 =
\bracks{\vec{p} - \pars{\vec{p} - \vec{p}\cdot\hat{n}_{-}\,\hat{n}_{-}}}\cdot
\bracks{\vec{p} - \pars{\vec{p} - \vec{p}\cdot\hat{n}_{+}\,\hat{n}_{+}}} =
\pars{\vec{p}\cdot\hat{n}_{-}}\pars{\vec{p}\cdot\hat{n}_{+}}
\hat{n}_{-}\cdot\hat{n}_{+}
$$
However, as we can see from expressions $\pars{1}$, 
$\ds{\vec{p}\cdot\hat{n}_{\pm} \not= 0}$ because $\ds{a^{2} \not= p^{2}}$ which leads to $\ds{\hat{n}_{-}\cdot\hat{n}_{+} = 0}$

Therefore, $\ds{\vec{p}}$ belong to the OP mentioned locus if there exists a pair of unit vectors $\ds{\hat{n}_{\pm}}$ such that:
\begin{equation}
\left\lbrace\begin{array}{rcl}
\ds{\pars{\vec{p}\cdot\hat{n}_{-}}^{2} + a^{2}} & \ds{=} & \ds{p^{2}}
\\
\ds{\pars{\vec{p}\cdot\hat{n}_{+}}^{2} + a^{2}} & \ds{=} & \ds{p^{2}}
\\
\ds{\hat{n}_{-}\cdot\hat{n}_{+}} & \ds{=} & \ds{0}
\end{array}\right.\tag{2}
\end{equation}
In writing $\ds{\vec{p}}$ as a linear combination of $\ds{\braces{\hat{n}_{\pm}}}$; namely,
$\ds{\vec{p} = c_{-}\hat{n}_{-} + c_{+}\hat{n}_{+}}$; we
find $\ds{c_{\pm}^{2} = a^{2}}$
$\ds{\pars{~\mbox{see conditions}\ \pars{2}~}}$ such that
$$
\color{#f00}{p} = \verts{\vec{p}} = \color{#f00}{\root{2}a}
$$


The locus is the set of points that rest in the circle
  $\color{#f00}{\ds{x^{2} + y^{2} = 2a^{2}}}$.

