Page:Elements of the Differential and Integral Calculus - Granville - Revised.djvu/204

 Substituting the values of $$x - \alpha$$ and $$y - \beta$$ from (D) in (A), and solving for $$R$$, we get

which is identical with (42), §103. Hence

Theorem. The radius of the circle of curvature equals the radius of curvature.



Here we shall make use of the definition of circle of curvature given in §104. Draw a figure showing the tangent line, circle of curvature, radius of curvature, and center of curvature $$(\alpha, \beta)$$ corresponding to the point $$P(x, y)$$ on the curve. Then

But $$BP = R \sin \tau, BC = R \cos \tau$$. Hence

From (29), §90, and (42), §103,

Substituting these back in (A), we get

From (23), §85, we know that at a point of inflection (as Q in the next figure)