<s>[INST] Polar Coordinates --- Equation of a line 
Hey, does anyone know how to tackle this question? I've tried using the formula r=d*sec(theta-alpha)but I'm not sure what each of the variables are equal to. If anyone can offer any help, it would be greatly appreciated!
 [/INST] 
The most direct approach, which would be preferred were this an exam question, is to find the equation for the line in Cartesian coordinates in the manner one would in pre-calculus, and then transform that result into polar coordinates.  The point $ \ (-11, -7) \ $ (which is on the circle of radius $ \ \sqrt{(-11)^2 + (-7)^2} \ = \ \sqrt{170} \ $ ) is on the radius with a slope of $ \ \frac{-7}{-11} \ \ , $ so the tangent line $  \ L \ $ has slope $ \ -\frac{11}{7} \ $ .
The equation of $ \ L \ $ is then
$$ \ y - (-7) \ = \ -\frac{11}{7} · (x - [-11] ) \ \ \Rightarrow \ \ 7y + 49 \ = \ -11x - 121 \ \ \Rightarrow \ \ 7y + 11x \ = \ -170 \ \ . $$
In polar coordinates, this becomes
$$ 7y + 11x \ = \ -170 \ \ \rightarrow \ \ 7r \ \sin \theta \ + \ 11r \ \cos \theta \ = \ -170 $$
$$ \Rightarrow \ \ r · (7  \ \sin \theta \ + \ 11  \ \cos \theta) \ = \ -170 \ \ \Rightarrow \ \  r \ = \ \frac{-170}{7   \sin \theta \ + \ 11   \cos \theta} \ \ . $$
The negative sign in the numerator of the expression places the tangency in the third quadrant.  (The solution given by Biswajit Banerjee works in an equivalent way, but the vector from $ \ O \ $ to $ \ P \ \ , \ \langle -11 \ , \ -7 \ \rangle \ , $ should have been used.)
As for the formula you inquired about, we could also start from the polar straight-line equation $ \ r \ = \ d · \sec(\theta - \alpha) \ \ , $ in which $ \ d \ $ represents the perpendicular (shortest) distance from the origin to the line and $ \ \alpha \ $ is the counter-clockwise angle by which the "vertical" line $ \ r \ = \ d · \sec \theta \ $ is rotated (making it also the angle that our line makes to the $ \ y-$axis.
Now this will mean that $ \ d \ = \ \sqrt{170} \ \ , $ which immediately raises the question of what happened to the radical and where the negative sign comes from.  Since $ \ d > 0 \ $ places $ \ r \ = \ d · \sec \theta \ $ in the half-plane "to the right" of the $ \ y-$axis , we will want to use $ \ d \ = \ -\sqrt{170} \ \ . $  Referring to the graph above, we can then express the rotation angle as $ \ \alpha \ = \ \arctan(7/11) \ = \ \arcsin(7/\sqrt{170}) \ = \ \arccos(11/\sqrt{170}) \ \approx \ 0.5667 \ \ . $  (Since $ \ 0 < \alpha < \frac{\pi}{2} \ $ , any of these is valid.)  So we could write the line equation as $ \ r  \ = \ -\sqrt{170} · \sec(\theta - \alpha) \ \ , $ applying the value determined.
Another way to work with the secant form is to invoke the angle-addition formula for cosine:
$$ r \ = \ d · \sec(\theta - \alpha) \ = \ \frac{d}{\cos(\theta - \alpha)} \ = \ \frac{d}{\cos \theta · \cos \alpha \  + \ \sin \theta · \sin \alpha }     \ \ . $$
For the moment, we'll pretend we don't know much about $ \ d \ $ or $ \ \alpha \ \ , $ other than that we'll choose to work with $ \ 0 < \alpha < \frac{\pi}{2} \ \ . $  We can use the Cartesian line equation or even just refer to the graph and make a similarity argument to find the intercepts of the line to be $ \ \left(-\frac{170}{11} \ , \ 0 \right) \ \ \text{and} \ \ \left(0 \ , \ -\frac{170}{7} \right) \ \ . $  Our polar equation then yields for the intercepts
$$ r(\pi) \ = \  \frac{d}{\cos \pi · \cos \alpha \  + \ \sin \pi · \sin \alpha} \ = \ \frac{d}{(-1) · \cos \alpha } \ = \ \ \frac{170}{11} $$
and
$$ r \left(\frac{3\pi}{2} \right) \ = \  \frac{d}{\cos \left(\frac{3\pi}{2} \right) · \cos \alpha \  + \ \sin \left(\frac{3\pi}{2} \right) · \sin \alpha} \ = \ \frac{d}{(-1) · \sin \alpha } \ = \ \ \frac{170}{7} \ \ , $$
since the intercepts lie at a positive radius from the origin in those directions.  We conclude that
$$   \cos \alpha  \ \ = \ \ -\frac{11}{170} · d \ \ , \ \ \sin \alpha  \ \ = \ \ -\frac{7}{170} · d $$
$$ \Rightarrow \ \ r \ =  \ \frac{d}{\cos \theta · \left(-\frac{11}{170} · d \right) \  + \ \sin \theta · \left(-\frac{7}{170} · d \right) } \ \ = \ \ \frac{-170}{11 \cos \theta   \  + \ 7 \sin \theta  } \ \ . $$
We asked earlier why there is no square-root in the numerator.  If we insert the actual trigonometric values for $ \ \alpha \ $ taken in the interval $ \ 0 < \alpha < \frac{\pi}{2} \   $ into the earlier version of the polar equation, we would have
$$   r \ =  \ \frac{-\sqrt{170}}{\cos \theta · \left(\frac{11}{\sqrt{170}}   \right) \  + \ \sin \theta · \left(\frac{7}{\sqrt{170}}   \right) }   \ \ , $$
which simplifies to our result.
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