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.

Given:

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.

Proof:

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,

ddx(x2+y2)=ddx(1)ddx(x2)+ddx(y2)=ddx(1)2x+ddx(y2)=0

Apply the chain rule and simplify the terms,

2x+[ddy(y2)dydx]=02x+[2ydydx]=02ydydx=−2xdydx=−xy

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.

m1m2=−x0y0y0x0=−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.