Question

The differential equation representing the family of circles with their centres on $x-$axis and whose radius is equal to the distance from from $(-1,2)$ to the line $3x+4y-15=0,$ is given by $y^2[{\left(\frac{dy}{dx}\right)}^2+k]=4,$ then $k^2+5$ is equal to

Answer 1

Radius $=\frac{|-3+8-15|}{\sqrt{3^2+4^2}}=\frac{10}{5}$
$\Rightarrow$ Radius $=2$
Now, equation of the family of circles having centres on $x-$axis and radius $2$ is
$(x-a{)}^2+(y-0{)}^2=2^2$
$\Rightarrow (x-a{)}^2+y^2=4\cdots (1)$
Diff. w.r.t $x,$ we get
$2(x-a)+2y\frac{dy}{dx}=0$
$\Rightarrow (x-a)=-y\frac{dy}{dx}\cdots \left(2\right)$
From $\left(1\right)$ and $\left(2\right),$ we have
$y^2{\left(\frac{dy}{dx}\right)}^2+y^2=4$
$\Rightarrow y^2[{\left(\frac{dy}{dx}\right)}^2+1]=4$
$\therefore k=1$
Hence, the value of $k^2+5$ is $6.$