Question

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

Short answer

The slope of the radius\(OP\)by using implicit differentiation is\(\frac{b}{a}.\)

Step-by-step solution

Given Information.

The tangent line at a point \(P\)to a circle with center \(O\) is perpendicular to the radius \(OP.\)

Meaning of implicit differentiation.

The procedure for determining the derivative of a dependent variable in an implicit function by differentiating each term independently, expressing the dependent variable's derivative as a symbol, and solving the resulting expression for the symbol.

Show that the tangent line at a point \(P\)to a circle with center \(O\) is perpendicular to the radius \(OP.\)

Assume that\(P(a,b)\)is a point on the circle\({x^2} + {y^2} = {a^2} + {b^2}\)

Differentiate both sides of the equation of the circle

\(\frac{{d\left( {{x^2} + {y^2}} \right)}}{{dx}} = \frac{{d\left( {{a^2} + {b^2}} \right)}}{{dx}}\)

The derivative in the right-hand side is\(0\)because that was for a constant.

\(\begin{array}{l}\frac{{d\left( {{x^2}} \right)}}{{dx}} + \frac{{d\left( {{y^2}} \right)}}{{dx}} = 0(SUMRULE)\\2{x^{2 - 1}} + \frac{{d\left( {{y^2}} \right)}}{{dy}} \cdot \frac{{dy}}{{dx}} = 0(CHAINRULE)\end{array}\)

Substitute the equation.

Subtract \(2x\)from both sides.

\(2{y^{2 - 1}} \cdot \frac{{dy}}{{dx}} = - 2x\)

Divide both sides by\(2y\)

\(\frac{{dy}}{{dx}} = - \frac{x}{y}\)

Substitute the point\(P(a,b)\)

\({\left. {\frac{{dy}}{{dx}}} \right|_{(a,b)}} = - \frac{a}{b}\)

Find the slope of the radius \(OP\).

\(\begin{array}{c}{m_1} = \frac{{b - 0}}{{a - 0}}\\ = \frac{b}{a}\end{array}\)