The curve for which the length of the normal is equal to the length of the radius vector, are

We have,

Length of the normal = Radius vector

$\begin{array}{l}\Rightarrow y\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}=\sqrt{x^2+y^2}\\ \Rightarrow y^2\left\{1+{\left(\frac{dy}{dx}\right)}^2\right\}=x^2+y^2\\ \Rightarrow y^2{\left(\frac{dy}{dx}\right)}^2=x^2\\ \Rightarrow x=\pm y\frac{dy}{dx}\Rightarrow x=y\frac{dy}{dx}\text{ or, }x=-y\frac{dy}{dx}\\ \Rightarrow xdx-ydy=0\text{ or, }xdx+ydy=0\\ \Rightarrow x^2-y^2=c_1\text{ or, }{x}^2+y^2=c_2\end{array}$

Clearly, $x^2-y^2=c_1$ represents a rectangular hyperbola and

$x^2+y^2=c_2$ represents circles.