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?

Here is a simple solution: Find the center of the given circle, which is (6, 8). The line from the center of the circle to the point of contact of one of the tangents is perpendicular to the tangent. Use this fact to get the equation of a new circle, which is $x^2-6x+y^2-8x=0$. Find where both circles meet. Find the angle between the vectors, which are the two points where the circles meet using the formula:$$\text{Angle} = \arccos\left(\frac{x_1x_2 + y_1y_2}{\sqrt{(x_1^2+y_1^2) \cdot (x_2^2+y_2^2)}}\right)$$
