The locus of centre of a circle which passes through the origin and cuts off a length of 4 units from the line x = 3 is 
 

$$\begin{gathered} (-\mathrm{g}, -\mathrm{f})=(\mathrm{x}, \mathrm{y}) be centre of circle \\ Given that 4=2\sqrt{{\mathrm{r}}^2-{\mathrm{d}}^2} \\ \Rightarrow {\mathrm{r}}^2-{\mathrm{d}}^2=4 \\ Given line x-3=0 \\ Now d=\frac{\left|-g-3\right|}{1} \\ \Rightarrow g^2+f^2-(g^2+9+6g)=4 \\ \Rightarrow f^2-6g=13 \\ \Rightarrow {(-f)}^2+6(-g)=13 \\ locus of (-g,-f) is y^2+6x=13 \\ {\mathrm{x}}^2+{\mathrm{y}}^2-{(\mathrm{x}-3)}^2=4 {\mathrm{x}}^2+{\mathrm{y}}^2-{\mathrm{x}}^2-9+6\mathrm{x}=4 \\ {\mathrm{y}}^2+6\mathrm{x}=13 \end{gathered}$$