Center on the x-axis, a distance ratio, and tangent slopes

Problem

In the coordinate plane, the center of circle $C$ lies on the $x$-axis, and the circle meets neither of the perpendicular lines

Let $d_1$ and $d_2$ be the shortest distances from points of the circle to $L_1$ and $L_2$. If $d_1$ is three times $d_2$, find the slope of a line through the origin that is tangent to the circle, in simplest radical form. The full question is on the CEEC 115 AST Mathematics I paper.

Where it goes wrong

Three stalls are common. One is to write “does not meet” as a distance greater than or equal to the radius: a tangent still meets the circle, so the distance must be strictly greater than the radius. Another is to use the distances from the center to the lines as $d_1$ and $d_2$. A third is to find the radius and then use the wrong condition for a tangent. The ratio fixes the circle first. The tangent slopes are the next step, not the same equation.

Idea

Let the center be $(a,0)$ and the radius be $r>0$. The distance from a point to a line gives

Missing both lines means ${\delta }_1>r$ and ${\delta }_2>r$. Then the shortest distances from the circle are

The condition $d_1=3d_2$ gives $r=|a|/2$. A line $y=mx$ through the origin is tangent when the distance from the center to that line equals the radius:

The slopes are $\pm \sqrt{3}/3$. The fill-in entry is written as $\pm$, $\sqrt{3}$, and $3$.

What the figure shows

• Drag a (x-coordinate of the center). The center moves along the $x$-axis. The radius is set to $|a|/2$, which makes $d_1:d_2=3:1$, and $a$ stays inside the range where the circle misses both lines.
• The short blue and purple segments are the shortest distances from the circle to $L_1$ and $L_2$. The ratio stays near $3:1$.
• The gold dashed lines are the two tangents through the origin. Their slopes are $\pm \sqrt{3}/3$, and those slopes do not change with $|a|$.