The curve for which the ratio of the length of the segment by any tangent on the $Y-$axis to the length of the radius vector is constant $(K)$, is

Slope of tangent at

$\begin{aligned} p(x, y) & \text { on } y=f(x) \\ = & \left.\frac{d y}{d x}\right|_{p(x, y)} \end{aligned}$

Equation of tangent at $P$

$\begin{array}{l} \Rightarrow \quad \frac{y-y}{x-x}=\frac{d y}{d x} \\ \Rightarrow \quad y-y=\frac{d y}{d x}(x-x) \end{array}$

for $y$-intercept put $x=0$,

$\gamma=y-x \frac{d y}{d x} \text { - (1) }$

Radius vector $\overline{O P}=\sqrt{x^{2}+y^{2}}$————-(2)

Given, $\quad \frac{y-x \frac{d y}{d x}}{\sqrt{x^{2}+y^{2}}}=k$

$\begin{aligned} & \sqrt{x^{2}+y^{2}} \\ \Rightarrow & y-x \frac{d y}{d x}=k\left(\sqrt{x^{2}+y^{2}}\right) \quad\left[\begin{array}{l} \text { Let } y=v x \\ \Rightarrow \frac{d y}{d x}=v+ \end{array}\right] \\ \Rightarrow & v x-x\left[v+x \frac{d v}{d x}\right]=k x \sqrt{1+v^{2}} \\ \Rightarrow & -x \frac{d v}{d x}=k \sqrt{1+v^{2}} \Rightarrow \int \frac{d v}{\sqrt{1+v^{2}}}=-k \int \frac{d x}{x} \\ \Rightarrow & \ln \left(v+\sqrt{1+v^{2}}\right)=-k \ln x+\ln c \\ \Rightarrow & v+\sqrt{1+v^{2}}=\frac{c}{x^{k}} \end{aligned}$

substituting $v=y / x$

$\begin{array}{l} \Rightarrow \frac{y}{x}+\sqrt{1+\left(\frac{y}{x}\right)^{2}}=\frac{c}{x^{k}} \\ \Rightarrow \frac{y+\sqrt{x^{2}+y^{2}}}{x}=\frac{c}{x^{k}}{ }^{k-1} \\ \Rightarrow\left(y+\sqrt{x^{2}+y^{2}}\right) x^{k-1}=c \end{array}$

$\therefore$ Hence, Option ( $B$ ) is the correct answer.