Question

Find the orthogonal trajectory of $x^2+y^2=c$

• A. $y=x^2$
• B. $x=y^2$
• C. $y=mx$
• D. $x^2-y^2=c$

Answer 1

The correct option is C

$y=mx$

Orthogonal trajectory of a family of curves is defined as the family of curves intersecting at right angles. We want to find the orthogonal trajectory of $x^2+y^2=c$. This circle has centre as the origin and variable radius. Since the centre is origin for all the circles in this family of curves, any line passing through origin will be a normal to all the circles. We can say that lines of the form $y=mx$ is the orthogonal trajectory of $x^2+y^2=c$.
There is a different method for finding orthogonal trajectory of any family of curves. Above method/explanation worked because we know the geometrical interpretation of the curves $x^2+y^2=c$.
Let’s look at the second method which would be more helpful in finding the orthogonal trajectory for a general curve
1. Find$\frac{dy}{dx}$ after differentiating the given equation and eliminate arbitrary constants
2. Replace $\frac{dy}{dx}\text{by}\frac{-dx}{dy}$
3. Solve the differential equation obtained in second step to find the orthogonal trajectory
Let's do the same for the equation given to us.
1. Differentiate $x^2+y^2=c$ to find $\frac{dy}{dx}$ and to eliminate $c$
we get $2x+2y\frac{dy}{dx}=0$
$\Rightarrow \frac{dy}{dx}=\frac{-x}{y}$
2. Replace $\frac{dy}{dx}\text{by}\frac{-dx}{dy}$
$\Rightarrow \frac{-dx}{dy}=\frac{-x}{y}$
$\Rightarrow \frac{dx}{dy}=\frac{x}{y}$
3. Solve above differential equation to obtain orthogonal trajectory
$\frac{dx}{x}=\frac{dy}{y}$
Integrating both the sides, we get
$\ln x=\ln y+k$
$\Rightarrow \ln \left(\frac{x}{y}\right)=k$
$\Rightarrow \frac{x}{y}=e^k$
$\Rightarrow y=e^{k}x$
This equation represents family of straight lines passing through origin as we inferred.
If we replace $e^k$ with $m$, we get $y=mx$