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: 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:$$\text{Angle} = \arccos\left(\frac{x_1x_2 + y_1y_2}{\sqrt{(x_1^2+y_1^2) \cdot (x_2^2+y_2^2)}}\right)$$
A: The mistake lies in using $6x+8y-75=0$ further instead of $-12x-16y+150=0$
in the formula. Correct way is to write:
$$C(0,0) C(x,y)=(-12x-16y+150)^2.$$
to get the correct answer.
A: Found the problem! To understand why it fails , refer back to the derivation here, it was assumed that:
$$C(P) + \lambda T(P)^2 = 0$$
Where we are subbing in the coordinate of point $P$ into equation of circle, and we assumed that $T(P) = C(P)$ , now this is not true if we directly scale up the equation of tangents (read Jocahimsthal's notation for understanding), now suppose we say,:
$$ T'(x,y) = \lambda T(x,y)$$
Then,
$$ \frac{T'(P)}{\phi} = T(x,y)$$
Or,
$$C(P) = \frac{T'(P)}{\phi} \tag{1} $$
And if we used scaled tangent equation in original equation
$$ C(P) +  \lambda (T'(P))^2=0 \tag{2}$$
From (1) and (2),
$$ C(P) + \lambda ( \phi C(P) )^2 = 0$$
Or,
$$ -\frac{1}{\phi^2 C(P)} = \lambda$$
Which changes our tangent equation to:
$$ C(x,y) C(P) = \frac{T'(x,y) )}{\phi^2}$$
A: Expansion on Math777's answer, see here first, the rest of the answer's idea can be understood in the following way:
Reference picture:

Fact: From any external point there exists two tangents to the circle, therefore there must be two tangents from the point $(0,0)$ to our circle since this point is external to our circle.
Now, the vector from origin to point of contact of tangent is perpendicular to radius vector because tangent and radius are perpendicular, there are two such points like this. These two points satisfy equation of the bigger diametric circle, hence these two points are the intersection of big diameteric circle and original one (See the linked post to understand the diameteric circle idea)
A: 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.
