Take a circle that is centered at the origin and two lines, 0.5 above and below the origin with slope $a$. The points that the lines intersect the circle to the right of the origin are named $A$ and $B$. What would be the function that takes the input of slope $a$ and returns the distance between the two points? It looks very similar to $1/n^{x^2}$, but I can't find any way to transform the function to fit the curve. I also tried using a CAS, but I couldn't get a function that fits the curve, and solving it by hand quickly proved to be extremely tedious. The values at $1$ and $-1$ are $1$ over the square root of $2$, and here is an image of the curve.

Curve image

  • $\begingroup$ Doesn't it depend on the radius of the circle? $\endgroup$ – Mark Fischler May 16 at 0:49
  • $\begingroup$ Nope, because any distance along the lines you go the distance between two points on them will not change $\endgroup$ – TigerGold May 16 at 0:52
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    $\begingroup$ Using brute-force elementary algebra (in Maple), I get a distance of $$\frac{1}{\sqrt{1+a^2}}$$ $\endgroup$ – quasi May 16 at 1:11
  • $\begingroup$ That works, I'm so stupid $\endgroup$ – TigerGold May 16 at 1:14
  • $\begingroup$ OMG, I guess we are all silly. The two lines are a fixed distance apart. By symmetry , the line between the intersection points with the circle is perpendicular to the two lines. So we can translate that line segment to touch $(0,\frac12)$ on one side. Then in the small right triangle formed by the line between $(0,-\frac12)$ and $(0,+\frac12)$, the perpendicular dropped to line A at point $P$, and the segment from the origin to P, the hypotenuse is $1$ and angle ABP is the same as the angle $\theta$ between the lines and the horizontal. So BP = $\cos \theta$ which is same as answer. $\endgroup$ – Mark Fischler May 16 at 15:12

Let the two points be at coordinates $(b_x,b_y)$ and $(a_x,a_y)$. Also, temporarily let the slope be called $s$, to avoid confusion with the coordinates of point $A$.
Then we have four equations in the four coordinates, saying that the two points lie on the circle and on the respective lines: $$ b_x^2 + b_y^2 = r^2 \\ a_x^2 + a_y^2 = r^2 \\ b_y =s b_x - \frac12 \\ a_y = s a_x + \frac12 $$ The solution to these equations with positive $a_x, b_x$ is $$ \pmatrix{a\\b} = \left( \frac{\sqrt{4(1+s^2)r^2-1} \mp s}{2(1+s^2)},\,\,\, \frac12\left( \pm \frac1{1+s^2} + \frac{\sqrt{4(1+s^2)r^2-1}}{1+s^2} \right)\right) $$ where the $+$ sign in $\pm$ refers to $a$ and the $-$ sign to $b$.

The answer $d$ is obtained by substituting these into $\sqrt{(x_x-a_x)^2+(b_y-a_y)^2}$ and simplifies to $$ d = \frac1{\sqrt{1+s^2}} $$ or in terms of $a$ as the slope,

$$ d = \frac1{\sqrt{1+a^2}} $$


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