Find the equation of circle concentric with the circle x^2 + y^2 + 4x – 2y + 4 = 0 and 3 times its radius 

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​$$\begin{gathered} \mathrm{We} \mathrm{have}: \\ {\mathrm{x}}^2+{\mathrm{y}}^2+4\mathrm{x}-2\mathrm{y}-4=0 \\ \mathrm{Center} \mathrm{of} \mathrm{this} \mathrm{cirlce}=\left(-\left(\frac{4}{2}\right),-\left(-\frac{2}{2}\right)\right)=\left(-2,1\right) \\ \mathrm{As} \mathrm{required} \mathrm{cirlce} \mathrm{is} \mathrm{concentric} \mathrm{with} \mathrm{above} \mathrm{cirlce} \mathrm{hence} \mathrm{centre} \mathrm{of} \mathrm{required} \mathrm{circle} \\ \mathrm{will} \mathrm{also} \mathrm{be} \left(-2,1\right) \\ \mathrm{Let} \mathrm{equation} \mathrm{of} \mathrm{required} \mathrm{cirlce} \mathrm{be}: \\ {\left(\mathrm{x}-\mathrm{h}\right)}^2+{\left(\mathrm{y}-\mathrm{k}\right)}^2={\mathrm{r}}^2, \mathrm{where} \left(\mathrm{h},\mathrm{k}\right) \mathrm{is} \mathrm{centre} \mathrm{and} \text{'}\mathrm{r}\text{'} \mathrm{is} \mathrm{radius}. \\ \Rightarrow {\left(\mathrm{x}+2\right)}^2+{\left(\mathrm{y}-1\right)}^2={\mathrm{r}}^2 \\ \mathrm{Radius} \mathrm{of} {\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}=0 \mathrm{is} \sqrt{{\mathrm{g}}^2+{\mathrm{f}}^2-\mathrm{c}} \\ \mathrm{Hence} \mathrm{radius} \mathrm{of} \mathrm{given} \mathrm{circle} {\mathrm{x}}^2+{\mathrm{y}}^2+4\mathrm{x}-2\mathrm{y}-4=0 \mathrm{is}: \\ \mathrm{R}=\sqrt{{\left(2\right)}^2+{\left(-1\right)}^2-\left(-4\right)}=\sqrt{4+1+4}=3 \\ \mathrm{Radius} \mathrm{of} \mathrm{required} \mathrm{cirlce} \left(\mathrm{r}\right)=3\mathrm{R}=3\times 3=9 \\ \mathrm{Hence} \mathrm{required} \mathrm{equation} \mathrm{of} \mathrm{circle} \mathrm{is}: \\ {\left(\mathrm{x}+2\right)}^2+{\left(\mathrm{y}-1\right)}^2=9^2 \\ \mathbf{\Rightarrow }{\left(\mathbf{x}\mathbf{+}\mathbf{2}\right)}^{\mathbf{2}}\mathbf{+}{\left(\mathbf{y}\mathbf{-}\mathbf{1}\right)}^{\mathbf{2}}\mathbf{=}\mathbf{81} \end{gathered}$$

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