WMC 2018 Senior Problem 14

Consider the above figure of the given situation. Note that $AC = 10$ and $BC = 3$ . By the Pythagorean theorem, $AB = \sqrt{10^2-3^2} = \sqrt{91}$ . Conveniently, the slope of a line is always the tangent of the angle it makes with the $x$ -axis (labeled $\theta$ in the figure), up to a factor of $-1$ . This is easy to calculate: $\tan\theta = \frac{AC}{AB} = \frac{3}{\sqrt{91}} = \frac{3\sqrt{91}}{91}$ Since point $B$ is in the third quadrant, the slope must be negative. Thus the answer is $\boxed{-\frac{3\sqrt{91}}{91}}$ .

Assume a line of slope m through (-10 , 0)

$y=m(x+10)$

Equate its perpendicular distance from origin to the circle's radius. (Making it a tangent)

$\frac {0-0+10m}{\sqrt {m^2+1}} = 3$

Solving this would give $m=+\frac {3}{√91} or -\frac {3}{√91}$ , but as the second point is in the third quadrant, the slope would be negative.