[sj18]

The question is as follows:

Recall that (x-h)^2 + (y-k)^2 = r^2. The circle's center is (h,k), with radius r.

Show that the tangent (derivative) to [(x-h)^2 + (y-k)^2 = r^2] is perpendicular to the radius at the point of tangency. Use implicit differentiation.

If anyone could help me with that question that would be great.

thanks.

P.S I need the answer by wednesdat January 18, 2007.

[Elisha Grey]

at a point (x,y) we have 2(x-h) + 2(y-k)der of y = 0, so der of y = -(x-h)/(y-k) = slope of a tangent line at this point. A radial line goes through (x,y) and (h,k), so its slope is (y-k)/(x-h), which is the negative reciprical of the slope of the tangent line. Hence these lines are perpendicular.