Find the curve whose curvature is 2, passes through the point (1,0) and whose tangent vector at (1,0) is [1/2 , (√3)/2 ].
The Attempt at a Solution
I know I must use the Fundamental theorem of plane curves but I don't know how to apply it correctly here. Another way to do it is the following but I don't know how to finish it:
Let us call our curve f(t). The curvature for f(t) is given by k=|f''(t)|=2. Since our curvature is a constant 2, we can think of this curve f(t) as a circle with radius r = 1/2. The equation for a circle is:
(x-a)^2 + (y-b)^2 = r^2, since the circle passes through (1,0) I can plug it in for (x,y) I get
(1-a)^2 + (0-b)^2 = (1/2)^2
(1-a)^2 + (b)^2 = 1/4
So we are given the points on the circle but we don't know what the center is. We are given the tangent vector.
So at (1,0) the tangent line is (1/2, (√3)/2). I can get the slope of this tangent line by rise over run (simple enough). Thus slope equals m = √3.
But how can I continue?