From point A(0,3) on the circle x^2 +4x+(y-3)^2 =0 a chord AB and extended to a point M such that AM=2AB.The equation of the locus of M

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$$\begin{gathered} \mathrm{We} \mathrm{have} \\ {\mathrm{x}}^2+4\mathrm{x}+{\left(\mathrm{y}-3\right)}^2=0 \\ \mathrm{Now} \\ \mathrm{A}\left(0,3\right) \mathrm{is} \mathrm{on} \mathrm{circle} \\ \mathrm{Let} \mathrm{point} \mathrm{M} \mathrm{be} \left(\mathrm{h},\mathrm{k}\right) \\ \mathrm{As} \\ \mathrm{AM}=2\mathrm{AB} \\ \Rightarrow \mathrm{AB}=\frac{\mathrm{AM}}{2} \\ \mathrm{Hence} \mathrm{B} \mathrm{is} \mathrm{mid} \mathrm{point} \mathrm{of} \mathrm{AM} \\ \mathrm{Coordinates} \mathrm{of} \mathrm{B} \mathrm{are} \left(\frac{\mathrm{h}+0}{2},\frac{\mathrm{k}+3}{2}\right)=\left(\frac{\mathrm{h}}{2},\frac{\mathrm{k}+3}{2}\right) \\ \mathrm{B} \mathrm{lies} \mathrm{on} \mathrm{cirlce} \\ \Rightarrow \frac{{\mathrm{h}}^2}{4}+4.\frac{\mathrm{h}}{2}+{\left(\frac{\mathrm{k}+3}{2}-3\right)}^2=0 \\ \Rightarrow \frac{{\mathrm{h}}^2}{4}+2\mathrm{h}+{\left(\frac{\mathrm{k}-3}{2}\right)}^2=0 \\ \Rightarrow {\mathrm{h}}^2+8\mathrm{h}+{\left(\mathrm{k}-3\right)}^2=0 \\ \mathrm{Hence} \mathrm{equation} \mathrm{of} \mathrm{locus} \mathrm{is} \\ {\mathrm{x}}^2+8\mathrm{x}+{\left(\mathrm{y}-3\right)}^2=0 \end{gathered}$$

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