#### To determine

**To show:** Any tangent line at the point *P* to the circle with center *O* perpendicular to the radius *OP*.

#### Explanation

**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.