Number of common tangents of circles x^2 + y^2 - 8x - 2y + 1=0 and x^2 + y^2 +6x +8y=0 are?

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$$\begin{gathered} \mathrm{We} \mathrm{have} \\ {\mathrm{x}}^2+{\mathrm{y}}^2-8\mathrm{x}-2\mathrm{y}+1=0 \\ \mathrm{Centre} \mathrm{is} {\mathrm{C}}_1\left(4,1\right) \\ \mathrm{Radius} ={\mathrm{r}}_1=\sqrt{4^2+1^2-1}=4 \\ {\mathrm{x}}^2+{\mathrm{y}}^2+6\mathrm{x}+8\mathrm{y}=0 \\ \mathrm{Centre} \mathrm{is} {\mathrm{C}}_2\left(-3,-4\right) \\ \mathrm{Radius}={\mathrm{r}}_2=\sqrt{{\left(-3\right)}^2+{\left(-4\right)}^2-0}=\sqrt{9+16}=\sqrt{25}=5 \\ {\mathrm{C}}_1{\mathrm{C}}_2=\sqrt{{\left(4+3\right)}^2+{\left(1+4\right)}^2}=\sqrt{49+25}=\sqrt{74} \\ \mathrm{Here} \\ {\mathrm{r}}_1+{\mathrm{r}}_2=4+5=9 \\ {\mathrm{r}}_2-{\mathrm{r}}_1=5-4=1 \\ \mathrm{As} \\ 1<\sqrt{74}<9 \\ \Rightarrow {\mathrm{r}}_2-{\mathrm{r}}_1<{\mathrm{C}}_1{\mathrm{C}}_2<{\mathrm{r}}_1+{\mathrm{r}}_2 \\ \mathrm{Hence} \mathrm{circle} \mathrm{cut} \mathrm{each} \mathrm{other} \mathrm{at} \mathrm{two} \mathrm{points}, \mathrm{hence} \mathrm{there} \mathrm{will} \mathrm{be} \mathrm{two} \\ \mathrm{common} \mathrm{tangents}. \end{gathered}$$

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