The locus of the foot of the perpendicular drawn from origin to a variable line passing through fixed point (2, 3) is a circle whose diameter is 
 

$$\begin{gathered} Let m be slope of line l=0 which is pas\sin g through the point (2,3) \\ Now equation of line be y-3=m(x-2) \\ \Rightarrow mx-y-2m+3=0\to \left(1\right) \\ let (h,k) be locus of foot of perpendicular from (0,0) to \left(1\right) \\ \frac{h-0}{m}=\frac{k-0}{-1}=\frac{-(0-0-2m+3)}{m^2+1} \\ \Rightarrow \frac{h}{m}=-k \\ \Rightarrow m=\frac{-h}{k} \\ from \left(1\right) \\ \frac{-h}{k}\left(x\right)-y-2\left(\frac{-h}{k}\right)+3=0 \\ \Rightarrow -hx-ky+2h+3k=0 \\ \Rightarrow locus of (h,k) is x^2+y^2-2x-3y=0 \\ \Rightarrow r=\sqrt{1+\frac{9}{4}}=\frac{\sqrt{13}}{2} \\ length of diametre=\sqrt{13} \end{gathered}$$