Finding the Polar Equation of a line 
Hello, this problem is giving me some trouble. I'm confused as to how I find the equation of the line. I know that x^2+y^2 = r^2, x=r$\cos\theta$, y=r$\sin\theta$, and that tan$\theta$= y/x. I'm just unsure as to how I can derive the polar equation.

The equation of the circle is $$x^2+y^2=370$$ Differentiating implicitly, we find $$2x+2y\frac{dy}{dx}=0$$ so $$\frac{dy}{dx}=-\frac{x}{y}$$So the slope of your line is $-\frac{19}{3}$. Therefore, in Cartesian coordinates, the slope of the line is $$y=-\frac{19}{3}(x+19)-3$$Now substitute $x=r\cos\theta$ and $y=r\sin\theta$ and solve for $r$.
Alternatively, if you don't want to use calculus, you can use a little elementary geometry: tangents are perpendicular to radii drawn to the point of tangency. Here, the slope of the line containing the radius from $(-19,-3)$ is obviously $\frac{3}{19}$, and perpendicular lines have slopes that are negative reciprocals, so the slope of the tangent line is $-\frac{19}{3}$.
