A curve passing through the point (1,1) has the property that the perpendicular distance of the origin from normal at any point P of the curve is equal to the distance of P from the x-axis is a circle with radius = _____

equation of normal at $\mathrm{P}\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)$; is

$\mathrm{y}-{\mathrm{y}}_1=-\frac{1}{\mathrm{m}}\left(\mathrm{x}-{\mathrm{x}}_1\right)$ $\mathrm{x}+{\mathrm{my}}_1-\left({\mathrm{x}}_1+{\mathrm{my}}_1\right)=0$ By the given data
$\frac{\left|{\mathrm{x}}_1+{\mathrm{my}}_1\right|}{\sqrt{1+{\mathrm{m}}^2}}={\mathrm{y}}_1;\mathrm{m}=\frac{{\mathrm{y}}_1^2-{\mathrm{x}}_1^2}{2{\mathrm{x}}_1{\mathrm{y}}_1};\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{{\mathrm{y}}^2-{\mathrm{x}}^2}{2\mathrm{xy}}$
Put y = vx and solving ${\mathrm{x}}^2+{\mathrm{y}}^2=\mathrm{cx}$
This passes through (1, 1) $\therefore$ ${\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}=0$
Which is a circle whose radius ; $\sqrt{1}=1$