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I want to construct two connecting arcs that have the same tangent at where they join. One has a constant radius of curvature and the other ends at a given point. To draw an example in your head:
What is the radius of curvature of the arc and where is it tangent to the circle? --Melab±1 ☎ 20:33, 8 May 2012 (UTC)
Let A be a given point outside a given circle, B the circle's center point and r the circle's radius. We're seeking a circular arc through A, which is internally tangent to the given circle and has radius R. Let C be the arc's centre and D an arc and circle meeting point. Then |CA| = |CD| = R and |CB| = R−r = |CA|−r. That means the arc's centre C is always closer by r to B than to A, so C is a point of one branch of a hyperbola of foci A,B, the arc's radius is R=|CA| and the tangency point D is found by intersecting the given circle with a |CB| line as the intersection which is not between B and C. The minimum R is (|AB|+r)/2.
The other branch of the hyperbola is a locus of centers of arcs through A which are externally tangent to the given circle.
If point A is inside the circle, then points C form an ellipse with foci A,B.
In the degenerate case of A being a point of the circle, the set of C's is a straight line AB and the arc sought is any circle tangent (internally or externally) to the given circle at A. --CiaPan (talk) 07:40, 11 May 2012 (UTC)