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. If the circle has radius r, its equation is x2 + y2 = r2 ⇒ + 2yy' = 0 ⇒ y' = , so the slope of the tangent line at P(x0, y0) is . The negative reciprocal of that slope is , which is the slope of OP, so the tangent line at P is perpendicular to the radius OP.

Answer 1

Answer:

The tangent line to the point P(x0,y0) and the line OP are perpendicular

Step-by-step explanation:

To show that the tangent line to the point P(x0,y0) is perpendicular to the line OP, of the circle with center at 0(0,0), you first use implicit differentiation on the equation of a circle, as follow:

$x^2+y^2=r^2$     (1)

equation of a circle with center at (0,0) and constant radius r.

The implicit derivative of the equation (1) is:

$\frac{d}{dx}(x^2+y^2)=\frac{d}{dx}r^2\\\\2x+2y\frac{dy}{dx}=0$

You solve the previous equation for dy/dt:

$\frac{dy}{dx}=-\frac{x}{y}$

The derivative dy/dx is also the slope of the tangent line, for the point P(x0,y0) you obtain:

$\frac{dy}{dx}=m=-\frac{x_o}{y_o}$

The slope m' of the OP line is given by:

$m'=\frac{y-0}{x-0}$

for the point P(x0,y0) you obtain:

$m'=\frac{y_0}{x_0}$

In order to know if the lines OP and tangent line to the point P, are perpendicular between them, the you verify the following relation:

$m'=-\frac{1}{m}$         (2)

In fact, you relace the values of m and m':

$\frac{y_o}{x_o}=-\frac{1}{x_o/y_o}\\\\\frac{y_o}{x_o}=\frac{y_o}{x_o}$

The tangent line to the point P(x0,y0) and the line OP are perpendicular