<s>[INST] 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?
 [/INST] 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.

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