# maximum points of intersection between a circle and rectangle

I had today this mathematical question:

What is the maximum number of points of intersection between a circle and a rectangle such that the length of the rectangle is greater than the circle's diameter, and its width is less than the circle's diameter?

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 I tried to fix the English -- I hope this reflects what you want to ask? – joriki Jun 22 '11 at 21:25 exactly yes sorry for bad writting – dato Jun 22 '11 at 21:31

Consider, for example, a circle of diameter 10 (radius 5) centered at the origin; hence its equation is $x^2 + y^2 = 25$. Consider a rectangle with vertices $(x_i, y_i)$ at $(-4, -8)$, $(4, -8)$, $(4, 4)$, $(-4, 4)$. Hence it's length (height) is $4 - (-8) = 12 > 10$, and its width is $4 - (-4) = 8 < 10$ (where 10 is the diameter of the circle). Then there are 2 points of intersection between the circle $x^2 + y^2 = 25$ and each of the line segments $y = 4$ ($-4 \leq x \leq 4$), $x = 4$ ($-8 \leq y \leq 4$), and $x = -4$ ($-8 \leq y \leq 4$), but no points of intersection between $x^2 + y^2 = 4$ and the rectangle's fourth side which lies on line $y = -8$ ($-4 \leq x \leq 4$). Solving for the points of intersection yields a total of 6 points of intersection of the circle and the rectangle: $(-4, -3), (-4, 3), (-3,4), (3, 4), (4, 3), (4, -3)$. ( If we move the circle vertically so it intersects the line $y = -8$, then it will no longer intersect the side along $y = 4$.