47. Show, using implicit differentiation, that any tangent line at a point $P$ to a circle with center $O$ is perpendicular to the radius $O P$.
To show: Any tangent line at the point P to the circle with center O perpendicular to the radius OP.
The center of the circle is O and radius of the circle is r.
Derivative rules: Chain rule
If y=f(u) and u=g(x) are both differentiable function, then dydx=dydu⋅dudx.
Consider the point Q is (0,0) and the point P is (x0,y0).
The equation of the circle is x2+y2=1 with center Q and radius r.
Differentiate x2+y2=1 implicitly with respect to x,
Apply the chain rule and simplify the terms,
Thus, the derivative of the equation is dydx=−xy.
That is, the slope of tangent to the equation is dydx=−xy.
Therefore, the slope of tangent to the equation at (x0,y0) is m1=−x0y0.
The slope of the line passing through the point (0,0) and (x0,y0) is m2=y0−0x0−0.
That is, the slope of the line (radius OP) is m2=y0x0.
Note that, if two lines are perpendicular, then the product of their slope is -1.
Therefore, any tangent line at the point P to the circle with center O perpendicular to the radius OP.
Hence the required proof is obtained.