The orthogonal trajectories of the family of curves y = cx^k are given by

x² + ky² = constant

Differentiating the given relation we have,

$\frac{dy}{dx}= ckx^{k-1} ⇒ c =\frac{1}{k} x^{1-k }\frac{dy}{dx}$

Putting this value of c in the given equation we have

$y = \frac{1}{k} x^{1-k }\frac{dy}{dx} x^k =\frac{1}{k} x \frac{dy}{dx}$

Replacing $\frac{dy}{dx}$ by $-\frac{dx}{dy}$, we get

$y = -\frac{1}{k} x\frac{dx}{dy}$ ⇒ ky dy + x dx = 0

$⇒ ky^2 + x^2$ = constant.

Hence (B) is the correct answer.