If the circle x² + y² + 4x + 2y - 4 = 0 is reflected in a mirror to become the circle x² + y² + 6x + 4y + 4 = 0, the mirror is lying along the straight line

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$$\begin{gathered} \mathrm{We} \mathrm{have} \\ {\mathrm{x}}^2+{\mathrm{y}}^2+4\mathrm{x}+2\mathrm{y}-4=0 \\ \mathrm{Centre} \mathrm{is} \left(-2,-1\right) \\ {\mathrm{x}}^2+{\mathrm{y}}^2+6\mathrm{x}+4\mathrm{y}+4=0 \\ \mathrm{Centre} \mathrm{is} \left(-3,-2\right) \\ \mathrm{Mid} \mathrm{point} \mathrm{of} \mathrm{two} \mathrm{centres} \mathrm{must} \mathrm{lie} \mathrm{on} \mathrm{mirror}. \\ \left(\frac{-2-3}{2},\frac{-1-2}{2}\right)=\left(-\frac{5}{2},-\frac{3}{2}\right) \\ \mathrm{Hence} \mathrm{equation} \mathrm{of} \mathrm{mirror} \mathrm{is} \\ \mathrm{y}+\frac{3}{2}=\mathrm{m}\left(\mathrm{x}+\frac{5}{2}\right) \\ \Rightarrow 2\mathrm{y}+3=\mathrm{m}\left(2\mathrm{x}+5\right) \\ \Rightarrow 2\mathrm{mx}-2\mathrm{y}+5\mathrm{m}-3=0 \\ \mathrm{Now} \mathrm{distance} \mathrm{of} \mathrm{line} \mathrm{from} \mathrm{centres} \mathrm{of} \mathrm{both} \mathrm{the} \mathrm{circle} \mathrm{must} \mathrm{be} \mathrm{same}. \\ \frac{2\mathrm{m}\left(-2\right)-2\left(-1\right)+5\mathrm{m}-3}{\sqrt{4{\mathrm{m}}^2+4}}=\frac{2\mathrm{m}\left(-3\right)-2\left(-2\right)+5\mathrm{m}-3}{\sqrt{4{\mathrm{m}}^2+4}} \\ \Rightarrow -4\mathrm{m}+2+5\mathrm{m}-3=-6\mathrm{m}+4+5\mathrm{m}-3 \\ \Rightarrow 2\mathrm{m}=2 \\ \Rightarrow \mathrm{m}=1 \\ \mathrm{Hence} \mathrm{mirror} \mathrm{is} \\ 2\mathrm{x}-2\mathrm{y}+2=0 \\ \mathrm{x}-\mathrm{y}+1=0 \end{gathered}$$

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