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.

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

Short answer

Differentiate the equation of a circle \({x^2} + {y^2} = ax\).

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

Differentiate the equation of a circle \({x^2} + {y^2} = by\).

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

Step-by-step solution

Differentiate the equations of the circles

Differentiate the equation of a circle \({x^2} + {y^2} = ax\).

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

Differentiate the equation of a circle \({x^2} + {y^2} = by\).

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

Find the equation of orthogonal trajectory 

The given curves are orthogonal at \(\left( {{x_0},{y_0}} \right)\). Therefore the slopes of the curves are defined for the following relation.

\(\begin{aligned}\frac{{a - 2{x_0}}}{{2{y_0}}} &= - \frac{{b - 2{y_0}}}{{2{x_0}}}\\2a{x_0} - 4x_0^2 &= 4y_0^2 - 2b{y_0}\\a{x_0} + b{y_0} &= 2\left( {x_0^2 + y_0^2} \right)\end{aligned}\)

Sketch the orthogonal trajectory

The circles \({x^2} + {y^2} = ax\) and \({x^2} + {y^2} = bx\) are intersecting at the origin where the tangents are vertical and horizontal.

The figure below represents the orthogonal trajectory of curves.