Prove that these perpendicular distances are in G.P. Let $BC$ be the chord of contact of the tangents from a point $A$ to the circle $x^2+y^2=1$. $P$ is any point on the arc $BC$. Let $PX, PY$ and $PZ$ be the lengths of the perpendiculars from P on the line $AB,  BC$ and $CA $ respectively then prove that $PX,  PY$,  and $PZ$ are $G. P.$
My approach : I considered point $A$ to be $(\alpha, \beta)$. Using this I got the equation of common chord as $$\alpha x+\beta y-1=0$$. Now I considered point P to be $(\gamma,  \delta)$.  For this circle the pair of tangents can be given as $$(x^2+y^2-1)({\alpha}^2+{\beta}^2-1)=(\alpha x +\beta y -1)^2$$ . Simplifying this we get $$(x-\alpha) ^2 + (y-\beta)^2=(y\alpha-x\beta)^2$$. Now I can get the perpendicular distances to the pair of tangents and the common chord from their equations but I am stuck in finding the perpendicular distance in the case of pair of tangents because I am not able to factorise it into pair of two straight lines nor do I know any formula to find the perpendicular distances without factorising the expression. Any other method to solve this problem is also appreciated.
 A: We may suppose $A(a,0)$ where $a\gt 1$.
Then, we have
$$B\left(\frac 1a,\frac{\sqrt{a^2-1}}{a}\right),C\left(\frac 1a,-\frac{\sqrt{a^2-1}}{a}\right)$$
So, the equation of the line $AB,BC,CA$ is given by
$$x+\sqrt{a^2-1}\ y-a=0,\qquad ax-1=0,\qquad x-\sqrt{a^2-1}\ y-a=0$$
respectively.
So, setting $P(\cos\theta,\sin\theta)$, we get
$$PX=\frac{|\cos\theta+\sqrt{a^2-1}\ \sin\theta-a|}{a}$$
$$PY=\frac{|a\cos\theta-1|}{a}$$
$$PZ=\frac{|\cos\theta-\sqrt{a^2-1}\ \sin\theta-a|}{a}$$
It follows from these that
$$\begin{align}PX\cdot PZ&=\frac{|\cos\theta+\sqrt{a^2-1}\ \sin\theta-a|}{a}\times \frac{|\cos\theta-\sqrt{a^2-1}\ \sin\theta-a|}{a}\\\\&=\frac{|(\cos\theta-a)^2-(a^2-1)\sin^2\theta|}{a^2}\\\\&=\frac{|\cos^2\theta+\sin^2\theta-2a\cos\theta-a^2(1-\cos^2\theta)+a^2|}{a^2}\\\\&=\frac{(a\cos\theta-1)^2}{a^2}=PY^2\end{align}$$
A: 
$$|\overline{PY}||\overline{PZ}| = 2r\sin^2\theta \cdot 2 r \sin^2\phi = \left(\;2r\sin\theta\sin\phi\;\right)^2 = |\overline{PX}|^2$$
A: I suspect there's an elegant synthetic solution to this, but after several hours of scribbling to no avail I resorted to trigonometry. My answer is very similar to mathlove's although I did derive it independently, and I think it's sufficiently different to be worth submitting.
WLOG, we can put $A$ on the +X axis. Let the angle that $AB$ makes with the -X axis be $\alpha$. Then
A is $(\sec\alpha, 0)$
B is $(\cos\alpha, \sin\alpha)$
C is $(\cos\alpha, -\sin\alpha)$  
The equations of the tangents can be written in normal form:
AB is $x\cos\alpha + y\sin\alpha - 1 = 0$
AC is $x\cos\alpha - y\sin\alpha - 1 = 0$  
Let P be $(\cos\theta, \sin\theta)$
Thus
$$\begin{align}\\ 
PY & = |\cos\theta - \cos\alpha|\\
PX & = |\cos\theta\cos\alpha + \sin\theta\sin\alpha - 1|\\
PZ & = |\cos\theta\cos\alpha - \sin\theta\sin\alpha - 1|\\
\end{align}$$
If $PX, PY, PZ$ are in geometric progression, $PX.PZ = PY^2$
$$\begin{align}\\
PX.PZ & = |(\cos\theta\cos\alpha - 1 + \sin\theta\sin\alpha)(\cos\theta\cos\alpha - 1 - \sin\theta\sin\alpha)|\\
& = |(\cos\theta\cos\alpha - 1)^2 - (\sin\theta\sin\alpha)^2\\
& = |(\cos\theta\cos\alpha - 1)^2 - (1 - \cos^2\theta)(1 - \cos^2\alpha)\\
& = |\cos^2\theta\cos^2\alpha - 2\cos\theta\cos\alpha + 1 - (1 - \cos^2\theta - \cos^2\alpha + \cos^2\theta\cos^2\alpha)|\\
& = |- 2\cos\theta\cos\alpha + \cos^2\theta + \cos^2\alpha|\\
& = (\cos\theta - \cos\alpha)^2\\
& = PY^2
\end{align}$$
A: To address your specific question, you can find the individual equations of the tangent lines via a procedure known as “splitting the conic.” If $C$ is the matrix of a degenerate conic that consists of two lines it will be some multiple of $gh^T+hg^T$, which has rank 2. Here $g$ and $h$ are the homogeneous vectors of the lines. The rank-1 matrices $gh^T$ and $hg^T$ generate the same set of points, but they’re not symmetric. If $p$ is the point of intersection of the two lines, then there’s some scalar $\lambda$ for which $C+\lambda[p]_\times$ generates the same conic as $C$, but has rank 1 ($[p]_\times$ stands for the “cross-product matrix” of $p$). The rows of this rank-1 matrix are multiples of one of the lines, while its columns are multiples of the other.  
Expanding and rearranging your equation for the lines gives $$(1-\beta^2)x^2+2\alpha\beta xy+(1-\alpha^2)y^2-2\alpha x-2\beta y+\alpha^2+\beta^2 = 0$$ which corresponds to the matrix $$C = \begin{bmatrix}1-\beta^2 & \alpha\beta & -\alpha \\ \alpha\beta & 1-\alpha^2 & -\beta \\ -\alpha & -\beta & \alpha^2+\beta^2\end{bmatrix}.$$ Their intersection has homogeneous coordinates $[\alpha:\beta:1]$, so $$C+\lambda[p]_\times = \begin{bmatrix} 1-\beta^2 & \alpha\beta-\lambda & \lambda\beta-\alpha \\ \alpha\beta+\lambda & 1-\alpha^2 & -\lambda\alpha-\beta \\ -\lambda\beta-\alpha & \lambda\alpha-\beta & \alpha^2+\beta^2 \end{bmatrix}.$$ This will have rank 1 if all of the $2\times2$ minors vanish. Taking the upper-left submatrix, we have $$(1-\alpha^2)(1-\beta^2)-(\alpha^2\beta^2-\lambda^2) = 0 \\ \lambda^2-(\alpha^2+\beta^2-1) = 0 \\ \lambda = \pm\sqrt{\alpha^2+\beta^2-1}.$$ $p$ is external to the circle, so $\alpha^2+\beta^2\gt1$ and $\lambda$ is real. The positive square root gives $$C+\lambda[p]_\times = \begin{bmatrix} 1-\beta^2 & \alpha\beta-\sqrt{\alpha^2+\beta^2-1} & -\alpha+\beta\sqrt{\alpha^2+\beta^2-1} \\ \alpha\beta-\sqrt{\alpha^2+\beta^2-1} & 1-\alpha^2 & -\beta-\alpha\sqrt{\alpha^2+\beta^2-1} \\ -\alpha-\beta\sqrt{\alpha^2+\beta^2-1} & -\beta+\alpha\sqrt{\alpha^2+\beta^2-1} & \alpha^2+\beta^2 \end{bmatrix}.$$ Taking the last row and column of this matrix, we get for the equations of the two tangent lines $$\left(\alpha\pm\beta\sqrt{\alpha^2+\beta^2-1}\right)\,x+\left(\beta\mp\alpha\sqrt{\alpha^2+\beta^2-1}\right)\,y = \alpha^2+\beta^2.$$

By the way, you can also use the cross product matrix together with the dual conic to produce the conic that represents the tangents through a point. For any two points $p$ and $q$, the homogeneous vector of the line through them is $g=p\times q=[p]_\times q$. For a nondegenerate conic $C$, $\det C\ne0$ and the dual conic is given by the matrix $C^{-1}$: if $h$ is a line tangent to the conic, then $h^TC^{-1}h=0$. Plugging $g$ into the equation, $$([p]_\times q)^TC^{-1}([p]_\times q) = q^T([p]_\times^T C^{-1} [p]_\times)\,q = 0,$$ i.e., the degenerate conic that consists of the tangents to $C$ through $p$ is $[p]_\times^T C^{-1} [p]_\times$. Using the unit circle $\operatorname{diag}(1,1,-1)$ and $p=[\alpha:\beta:1]$ in this formula produces the same matrix as above.
