Tangent Secant Theorem/Proof 2

Proof
Let the circle $\CC$ be embedded in a Cartesian plane with its center located at the origin.

Let $\CC$ have radius $r$.

From Equation of Circle center Origin, $\CC$ can be described as:
 * $(1): \quad x^2 + y^2 = r^2$

From Equation of Straight Line in Plane: Parametric Form, let the straight line $DCA$ be described using the parametric equations:


 * $\begin {cases} x = x_0 + \rho \cos \psi \\ y = y_0 + \rho \sin \psi \end {cases}$

where $D = \tuple {x_0, y_0}$.

Substituting these for $x$ and $y$ into equation $(1)$:

We see that $(2)$ is a quadratic equation in $\rho$.

Let $\rho_1$ and $\rho_2$ be the solutions of $(2)$.

Each of these corresponds to one of the points at which $DCA$ intersects $\CC$, where:
 * $\rho_1 = DA$
 * $\rho_2 = DC$

From Product of Roots of Quadratic Equation:
 * $\rho_1 \rho_2 = x_0^2 + y_0^2 - r^2$

From Length of Tangent from Point to Circle center Origin:
 * $DB = x_0^2 + y_0^2 - r^2$

Hence we have:
 * $DA \cdot DC = DB^2$