b(s)=(4/5 Cos(s),1- Sin(s),-3/5 Cos(s))

is this curve is a circle and find the center and the radius ?

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- Dec 4th 2010, 12:46 AMnice rosehow to know the circle

b(s)=(4/5 Cos(s),1- Sin(s),-3/5 Cos(s))

is this curve is a circle and find the center and the radius ?

- Dec 4th 2010, 02:12 AMemakarov
This parametric function has a period $\displaystyle 2\pi$, so it seems that b(0) and b($\displaystyle \pi$), as well as b($\displaystyle \pi/2$) and b($\displaystyle 3\pi/2$) should be at the opposite points of the circle, if it is indeed a circle. So the center should be in the middle of the segment [b(0), b($\displaystyle \pi$)]. I got (0, 1, 0) as the center coordinates. So far it's a guess that this is a circle and that this is indeed the center.

One can check that b(s) is equidistant from (0, 1, 0). So one is only left to check that the figure lies in one plane. The first thing that came to mind is to take two vectors: from (0, 1, 0) to b(0) and from (0, 1, 0) to b($\displaystyle 3\pi/2$) and to compute a perpendicular to them using the dot product. This implies that the perpendicular vector (x, y, z) is subject to y = 0 and 4x - 3z = 0. E.g., (3, 0, 4) is one perpendicular vector. Then one can check that any vector from (0, 1, 0) to b(s) is perpendicular to (3, 0, 4), so all b(s) indeed lie in one plane.