Circles - point of intersection of tangents Question:
Let $A$ be the center of the cricle $x^2 + y^2 - 2x-4y-20=0$. Suppose that the tangents at the points $B(1,7)$ and $D(4,-2)$ on the cricle meet at point $C$. Find the area of the quadrilateral $ABCD$.
What I have done:
Well I have found the center of the circle and its radius. Upon drawing the diagram, it is obvious that the quadrilateral formed can be split into two right angled triangles. The only thing I need is the distance between the point of contact and the point of intersection of the tangents. How would I obtain this?
 A: Hint.  You know the lengths $AB$ and $AD$, and you know that $\angle ABC$ and $\angle ADC$ are right angles.  Now find $\angle CAB=\frac{1}{2}\angle DAB$.

Alternative.  The radius $AB$ is vertical, so the tangent $BC$ is horizontal, so $C$ has coordinates $(1+x,7)$ for some $x$.  The area will be $5x$.  Can you see how to find $x$ by using the fact that distances $BC$ and $DC$ are equal?
A: The only thing you need is the coordinate of $C$. If you think of the quadric as that of a polarity, you probably realize $BD$ the polar line of the pole $C$. In this way, you can quickly get the coordinate of $C$. 
More specifically, since $C$ is a point outside your circle, there are two lines that are tangent to the circle and also pass through $C$. The polar line to $C$ is given by the span of the two points of tangency, which in you case is just the line passing through $B$ and $D$. This is more of a projective geometry concept. But anyway, a formula is readily available for you, see here. I don't need to type it here do I? So first, you write down the line equation of $BD$, and you apply the formula. And you get your coordinate for $C$. 
A: Find the distance $BD$ using the coordinates of $B$ and $D$. Let $E$ be the midpoint of $BD$,
so $BE = \frac12 BD$. Observe that $\triangle AEB$ is a right triangle.
Knowing $BD$ and $AB$, apply the Pythagorean formula to find $AE$.
You can now use similar triangles to show that $BE:AE = BC:AB$.
Solve for $BC$ in that last equation.
A: The center of the circle is quickly found by derivatives of the equation
$$ \left. \begin{aligned} 
  \frac{{\rm d}}{{\rm d}x} (x^2+y^2-2x -4 y-20) &= 0 \\
  \frac{{\rm d}}{{\rm d}y} (x^2+y^2-2x -4 y-20) &= 0 
\end{aligned} \right\} 
\begin{aligned} x &= 1 \\ y &= 2 \end{aligned} $$
I like to use homogeneous coordinates, and so $A=(1,2,1)$ is the center of the circle. We also know $B=(1,7,1)$ and $D=(4,-2,1)$.
The tangent lines through B and D are found by 
$$ \begin{aligned}
  L_B &= U\, B = (0,5,-35) \} 0x+5y-35 =0 \\ 
  L_D & = U\, D =(3,-4,-20) \} 3x-4y-20=0 \\ 
  U & = \begin{bmatrix} 1 & 0 & -x_A \\ 0 & 1 & -y_A \\ -x_A & -y_A & -x_A^2-y_A^2-r^2 \end{bmatrix} 
=\begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & -2 \\ -1 & -2 & -20 \end{bmatrix} 
\end{aligned} $$ 
where $U$ is a 3×3 matrix representing the circle such that the equation of the circle is found by the quadratic form 
$$ \begin{pmatrix} x & y & 1 \end{pmatrix} \begin{bmatrix} 1 & 0 & -1 \\ 0 & 1 & -2 \\ -1 & -2 & -20 \end{bmatrix} \begin{pmatrix} x \\ y \\ 1 \end{pmatrix} =0 \} x^2+y^2 -2 x - 4 y -20 =0$$
Point $C$ is found where $L_A$ and $L_B$ meet 
$$ \left. \begin{aligned}
  C & = L_A \times L_B \\
    & = (0,5,-35) \times (3,-4,-20) = (-240,-105,-15)
\end{aligned} \right\} \begin{aligned} x_C & = 16 \\ y_C & = 7 \end{aligned} $$
Now the area of the triangle $$\triangle_{BAD} = \frac{1}{2} \frac{ A \cdot (B \times D) }{|A| |B| |D|} = \frac{15}{2}$$
where $\cdot$ is the inner product, $\times$ is the cross product and $|(a,b,c)|=c$ correspond to the scalar component of the homogeneous coordinates.
Also the area of the triangle $$\triangle_{DCB} = \frac{1}{2} \frac{ C \cdot (D \times B) }{|C| |D| |B|} = \frac{135}{2}$$
Combined you have $\boxed{\triangle_{BAD}+\triangle_{DCB} = 75}$
The beauty of homogeneous coordinates is that a) no trigonometry is needed and b) if the coordinates are rational then the result is a rational number. So integer algebra is sufficient to produce results.

