Question

Let differential equation of family of circles touching $y-$axis at the origin be $\frac{dy}{dx}+\frac{x^2-\lambda y^2}{\mu xy}=0$, where $\lambda \in R$, $\mu \in R-\left\{0\right\}$. Then the value of $\lambda +\mu$ is

Answer 1

Equation of family of circles touching $y-$axis at the origin is
$(x-a{)}^2+y^2=a^2$ where $a$ is radius of the circle.
$\Rightarrow x^2+y^2=2ax\cdots \left(i\right)$
Differentiating w.r.t. $x$, we get
$2x+2y{y}^{\prime }=2a$
$\Rightarrow x+y{y}^{\prime }=a\cdots \left(ii\right)$
Dividing $\left(i\right)$ and $\left(ii\right),$ we get
$\frac{x^2+y^2}{x+y{y}^{\prime }}=\frac{2x}{1}$
$\Rightarrow x^2+y^2=2x^2+2xy{y}^{\prime }$
$\Rightarrow y^2-x^2=2xy{y}^{\prime }$
$\Rightarrow \frac{dy}{dx}+\frac{x^2-y^2}{2xy}=0$
$\therefore \lambda =1,\mu =2$
Hence, the value of $\lambda +\mu$ is $3.$