Find the number of common tangents to the circles $x^2+y^2-8x+2y=0$ and $x^2+y^2-2x-16y+25=0$.

 

Let the first circle be $C_1=x^2+y^2-8x+2y$

$$\begin{gathered} x^2+y^2-8x+2y=0 \\ {x}^2-2\times 4\times x+16+y^2+2\times y+1-16-1=0 \\ {\left(x-4\right)}^2+{\left(y+1\right)}^2=17 \\ {\left(x-4\right)}^2+{\left(y+1\right)}^2={\left(\sqrt{17}\right)}^2 \end{gathered}$$

The center of this circle $C_1$is $\left(4,-1\right)$ and radius $r_1=\sqrt{17}$units

And, let the second circle $\left(C_2\right)$ be

$$\begin{gathered} x^2+y^2-2x-16y+25=0 \\ \Rightarrow x^2-2\times x+1+y^2-2\times 8y+64-1-64+25=0 \\ \Rightarrow {\left(x-1\right)}^2+{\left(y-8\right)}^2=40 \\ \Rightarrow {\left(x-1\right)}^2+{\left(y-8\right)}^2={\left(\sqrt{40}\right)}^2 \end{gathered}$$

The center of this circle $C_2$ is $\left(1,8\right)$ and radius $r_2=\sqrt{40}$units.

Now, Distance between the center of the two circles 

$$\begin{gathered} =C_1{C}_2 \\ =\sqrt{{\left(4-1\right)}^2+{\left(-1-8\right)}^2} \\ =\sqrt{90} \\ =9.49 \end{gathered}$$

Sum of the radius of the circles 

$$\begin{gathered} =\sqrt{40}+\sqrt{17} \\ =6.32+4.12 \\ =10.44 \end{gathered}$$

Since, the distance between the center of the circles is less than the sum of their radius

Therefore, the circles intersect each other

Therefore, the number of common tangents for both the circles will be two.

Hence, the correct answer is $2$.