Find nature of curve not passing through origin for which tangent at any point P. on curve is equal to radius vector of point P.

$(a)\;x^2-y^2=c \\ (b)\;y^2+9x^2+6=c \\ (c)\;x^2+3xy^2-y=c \\ (d)\;x^2+3y^2=c$

Equation of normal at P
$Y-y=-\large\frac{-1}{m}$$(X-x) => X+mY=x +my PN=\sqrt {m^2y^2+y^2} OP= PN x^2+y^2=m^2y^2+y^2 x^2 =m^2y^2 \large\frac{dy}{dx}$$=m =\pm \large\frac{x}{y}$
=> $\int y dy =\pm \int x dy$
$x^2+y^2=c$ (circle)
or $x^2-y^2=c$
Hence a is the correct answer.