Question

Show, using implicit differentiation, that any tangent line at a point P to a circle with center O is perpendicular to the radius OP. I can visualize it/draw a diagram but the actual calculations aren't quite clear to me still =\

Answer 1

The slope at any point (x,y) on the circle is dy/dx.
For a circle with radius r, using implicit differentiation, we have:
\[{d\over dx}(x^2+y^2)={d \over dx}(r^2)\]
Since y=f(x), we use the chain rule on y^2:
\[2x+2yy'=0\\
\Rightarrow y'=-{x \over y}\]
This means that a tangent line on the point (x1,y1) on the circle has slope -x1/y1.
The slope of a line through the origin passing through the point (x1,y1) is y1/x1.
Thus these two lines are perpendicular.