Question

53-56 Two curves are orthogonal if their tangent lines are perpendicular at each point of intersection. Show that given families of curves are orthogonal trajectories of each other; that is, every curve in one family is orthogonal to every curve in the other family. Sketch both families of curves on the same axes.

53. \({x^{\bf{2}}} + {y^{\bf{2}}} = {r^{\bf{2}}}\), \(ax + by = {\bf{0}}\)

Short answer

Let the equation of the circle with centre at the origin \(\left( O \right)\)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.

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

Step-by-step solution

Find the slope of the tangent to the circle

Let the equation of the circle with centre at the origin \(\left( O \right)\)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.

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

Find the slope of the line \(ax + by = {\bf{0}}\)

The line \(ax + by = 0\) is passing through the origin.

Differentiate the equation \(ax + by = 0\).

\(\begin{aligned}a + by' &= 0\\y' &= - \frac{a}{b}\end{aligned}\)

By the equation of line \(ax + by = 0\).

\(\begin{aligned}{c}by &= - ax\\\frac{y}{x} = - \frac{a}{b}\end{aligned}\)

Therefore, the slope of the line is \(\frac{y}{x}\).

Sketch the orthogonal trajectory

As the slope of the line at \(\frac{y}{x}\), and the slope of the tangent \( - \frac{x}{y}\), which is negative reciprocal. Thereforetangent to the circle and the line \(ax + by\) are perpendicular to each other. So, the curves are orthogonal, and the family of the curves are orthogonal trajectories to each other.

The figure below represents the orthogonal trajectory of curves.