I'm not sure best how to describe this problem, but here it is:

I need to find an equation describing the midpoints of a rectangle bounding a circle: The equation for the circle is:

$$(x-200)^2 + (y-200)^2 = 200^2$$

The lines describing the rectangle are:

$$\begin{align} x&=0\\ x&=800\\ y&=0\\ y&=600\\ \end{align}$$

I need an equation describing the midpoints between any lines perpendicular to the gradient of the circle, and the rectangle boundary it touches. Any help would be appreciated.

low quality :(

  • 1
    $\begingroup$ "the midpoints between any lines perpendicular to the gradient of the circle, and the rectangle boundary it touches" -- I'm having a hard time interpreting this description. Could you add a sketch illustrating what you mean? $\endgroup$ – Rahul May 13 '14 at 7:24
  • $\begingroup$ here is my low quality illustration - I think you get the idea $\endgroup$ – duncan mckellar May 18 '14 at 7:23

Here's a simplified, generic version of the problem.

Center a circle of radius $r$ a the origin, and consider the movable point $A$ on line $x=a$. Let $\alpha$ be the angle that $\overline{OA}$ makes with the $x$-axis, and let $A^\prime$ be the point where $\overline{OA}$ meets the circle.

enter image description here

Then we can write $$A = (a, a\tan\alpha) \qquad A^\prime = ( r \cos\alpha, r \sin\alpha )$$ so that the midpoint of $\overline{AA^\prime}$ is $$M = \frac12(A+A^\prime) = \frac{1}{2}\left(\; a + r \cos\alpha, \; a \tan\alpha + r \sin\alpha \;\right)$$ Likewise, we have $$N = \frac{1}{2}\left(\; b \tan\beta + r\sin\beta,\; b + r\cos\beta\;\right)$$

I'm unaware of any specific name given to this locus of midpoints. However, the curve resembles the Witch of Maria Agnesi.

In your case, we have $r = 200$ and center the circle at $O(r,r)$. Since $a$ and $b$ are the distances from the circle's center to the lines, we also have $a = 800-r = 600$ and $b = 600 - r = 400$. Finally, we'll need to translate the points $M$ and $N$ upward and rightward by $r$:

$$\begin{align} M &= (200,200) + \frac{1}{2}\left(\;600 + 200\cos\alpha,\;600\tan\alpha + 200\sin\alpha\;\right) \\ &= \left(\;500+100\cos\alpha,\;200+300\tan\alpha+100\sin\alpha\;\right) \\[4pt] N &= \left(\;200+200\tan\beta+100\sin\beta,\;400+100\cos\beta\;\right) \end{align}$$

Presumably, you want to jump from the $M$ curve to the $N$ curve at the point where they intersect; this point corresponds to $A$ and $B$ coinciding at $Z$, so that $$\tan\alpha = \frac{b}{a} =\frac{1}{\tan\beta}$$

Finding the formulas for the counterpart curves for the other two sides of the bounding rectangle is left as an exercise to the reader.

  • $\begingroup$ thanks, here it is: c = 0 - r = -200 and d = 0 - r = -200 O = (200,200) + 0.5 (-200 + 200cos(delta), -200tan(delta) + 200sin(delta) = (100 + 100cos(delta), 200 - 100tan(delta) + 100sin(delta)) P = (-100 + 100 tan(gamma) + 100cos(gamma), -100 + 100 cos(gamma)) $\endgroup$ – duncan mckellar May 20 '14 at 13:38

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