I recently came across a problem that asked me to find the minimum value of $\frac{j}{k}$ such that the imaginary number $(j - 5) + ki$ has modulus equal to 2 (and such that $j$ and $k$ are both real numbers).
My current idea to solve this problem is just to utilize the modulus. Essentially, $|(j-5) + ki| = \sqrt{(j-5)^2+k^2} = 2$, so $(j-5)^2 + k^2 = 4$. Then, I realized that this equation represents a circle on the coordinate plane, so the maximum value of $\frac{j}{k}$ is just equal to the maximum slope of any line passing through the origin that hits the circle, or the positive slope of the tangent line from the origin to the circle.
From this, I know one way to solve this problem, but I feel that this is more tedious than I expected. Can someone point me to any simpler way to solve this problem? Is there another geometric analogy that I am missing here?
I think you're almost there. Draw three points: origin, the center of the circle and the point at which the straight line tangents the circle. From there, you can use the trigonometry of right triangles to say that the angle the line forms to the positive $x$-axis (lying below it) obeys
$$\sin \theta =2/5$$.
Then, since the slope of a line is the tangent of the angle it forms to the positive $x$-axis, we have
$$\frac{j}{k} = - \tan \sin^{-1} 2/5 = -\frac{2}{\sqrt{21}}$$.