From the point A(0, 3) on the circle $x^2+4x+{\left(y-3\right)}^2=0$ a chord AB is drawn and extended to a point P such that AP = 2AB then the locus of P is 
 

$$\begin{gathered} \mathrm{Given} \mathrm{equation} \mathrm{of} \mathrm{circle} \mathrm{is} {\mathrm{x}}^2+4\mathrm{x}+{(\mathrm{y}-3)}^2=0,\mathrm{A}=(0,3) \\ \mathrm{Let} \mathrm{P}=(\mathrm{h},\mathrm{k}) \mathrm{be} \mathrm{locus} \mathrm{of} \mathrm{point} \\ \mathrm{and} \mathrm{given} \mathrm{AP}=2\mathrm{AB} \Rightarrow \mathrm{B} \mathrm{is} \mathrm{the} \mathrm{mid}-\mathrm{point} \mathrm{of} \mathrm{AP} \\ \Rightarrow \mathrm{B}=(\frac{\mathrm{h}}{2},\frac{\mathrm{k}+3}{2}) \\ \mathrm{It} \mathrm{lie} \mathrm{on} \mathrm{the} \mathrm{circle} {\mathrm{x}}^2+4\mathrm{x}+{(\mathrm{y}-3)}^2=0 \\ \mathrm{sub} \mathrm{x}=\frac{\mathrm{h}}{2},\mathrm{y}=\frac{\mathrm{k}+3}{2} \\ \therefore \frac{{\mathrm{h}}^2}{4}+2\mathrm{h}+{(\frac{\mathrm{k}+3}{2}-3)}^2=0 \\ \Rightarrow \frac{{\mathrm{h}}^2}{4}+2\mathrm{h}+\frac{{\mathrm{k}}^2-6\mathrm{k}+9}{4}=0 \\ \therefore {\mathrm{h}}^2+{\mathrm{k}}^2+8\mathrm{h}-6\mathrm{k}+9=0 \\ \therefore \mathrm{locus} \mathrm{of} \mathrm{P} \mathrm{is} {\mathrm{x}}^2+{\mathrm{y}}^2+8\mathrm{x}-6\mathrm{y}+9=0 \end{gathered}$$