Question

Show using implicit differentiation that any tangent line at a point P to a circle with centre O is perpendicular to the radius\(OP\).

Short answer

The product of slopes of tangent and radius is \( - 1\); therefore, radius and tangent are perpendicular.

Step-by-step solution

Find the slope of the tangent to the circle

Let the equation of the circle with centre at the origin be:

\({x^2} + {y^2} = {r^2}\)

Here, r is the radius of the circle.

Differentiate the equation of the circle to find the slope of its tangent.

\(\begin{array}{c}2x + 2yy' = 0\\y' = - \frac{x}{y}\end{array}\)

The slope of the tangent at the point \(\left( {{x_0},{y_0}} \right)\) is:

\(y'\left( {{x_0},{y_0}} \right) = - \frac{{{x_0}}}{{{y_0}}}\)

Find the relation between the slope of tangent and radius

The radius of the circle is given by the line joining the points \(\left( {0,0} \right)\) and \(\left( {{x_0},{y_0}} \right)\). The slope of the radius is:

\(\begin{array}{c}{m_r} = \frac{{{y_0} - 0}}{{{x_0} - 0}}\\ = \frac{{{y_0}}}{{{x_0}}}\end{array}\)

Find the product of slope of tangent and the radius of the circle:

\( - \frac{{{x_0}}}{{{y_0}}} \times \frac{{{y_0}}}{{{x_0}}} = - 1\)

The figure below represents the circle and tangent to it.

As the product of slopes of tangent and the radius is \( - 1\), tangent and radius are perpendicular to each other.