Angle between tangents to circle given by $x^2 + y^2 -12x -16y+75=0$? 
Given the circle: $C(x,y)=x^2 + y^2 -12x -16y+75=0$, find the two tangents from origin

First, I get the line which passes through point of contact of tangents from origin using result here which is :
$$ -12x-16y-2 \cdot 75 = 0$$
Or,
$$ 6x + 8y -75 =0 $$
Now, I use the result discussed in this answer, which says that pair of straight from point $(P)$ to conic is given as:
$$ C(0,0) C(x,y) =  (6x+8y-75)^2$$
This leads to:
$$ 75 (x^2 + y^2 -12x-16y+75) = (6x+8y-75)^2$$
$$ 0 = 75(x^2 +y^2 - 12 x - 16y +75) - (6x+8y-75)^2$$
For applying the result in this answer, then the formula
$$ a= 75 - 36,  b=75-64 , h= - \frac{(6 \cdot 8 \cdot 2 )}{2}$$
Or,
$$ a = 39, b= 11 , h=-48$$
$$ \tan \theta = \frac{2 \sqrt{(-48)^2-39 \cdot 11)}}{39+11} = \frac{2 \sqrt{5^4 \cdot 3}}{25} = 2 \sqrt{3}$$
However, the intended answer was:
$$ \tan \theta = \frac{1}{\sqrt{3} } $$
Where have I gone wrong?

A much more sane and simpler way:

Observe the right triangle created by the line segment from origin to center, and origin to the point which is tangent to circle. Since, tangent and radius are perpendicular, we can apply good ol' trignometery easily:
$$ \sin \theta = \frac{r}{OC}= \frac{5}{10}= \frac{1}{2}$$
Hence, the $ x= \frac{\pi}{6}$ , which makes total angle between tangent as $ x = \frac{\pi}{3}$
However, all the other answer are beautiful in demonstrating the different way of approaching the same problem.
