The least distance from origin to a point on the line y = x + 3 which lies at a distance of 2 units from 0 ,…

Let the point be $\left(x,y\right)$ on the line $y=x+3$,

Now slope of the line is $m=1\Rightarrow \theta =45°$,

Now given point is at $2$ unit distance from point $\left(0,3\right)$

Now by parametric form we get,

$\frac{x-0}{\cos \theta }=\frac{{\displaystyle y-3}}{\sin {\displaystyle \theta }}=\pm 2$

$\Rightarrow \frac{x-0}{\frac{1}{\sqrt{2}}}=\frac{{\displaystyle y-3}}{\frac{1}{\sqrt{2}}}=\pm 2$ $\left\{\text{as }\theta =45°\right\}$

$\Rightarrow x=\pm \sqrt{2} \& y=\pm \sqrt{2}+3$

Now distance from origin is given by $D=\sqrt{{\left(\pm \sqrt{2}-0\right)}^2+{\left(\pm \sqrt{2}+3-0\right)}^2}$

Taking positive sign we get, $D=\sqrt{13+6\sqrt{2}}$

And taking negative sign we get, $D=\sqrt{13-6\sqrt{2}}$

So least distance is $D=\sqrt{13-6\sqrt{2}}$