The number of common tangents of the circles given by $$x^2 + y^2 - 8x - 2y + 1 = 0$$ and $$x^2 + y^2 + 6x + 8y = 0$$ is

The general equation of a circle $$x^2+y^2+2gx+2fy+c=0$$ has centre $$(-g,-f)$$ and radius $$r=\sqrt{g^{\,2}+f^{\,2}-c}$$. We first convert both equations to their standard forms.

Circle 1
$$x^2+y^2-8x-2y+1=0$$
Group the $$x$$ and $$y$$ terms:

$$\bigl(x^2-8x\bigr)+\bigl(y^2-2y\bigr)= -1$$

Complete the squares:
$$\bigl(x-4\bigr)^2-16+\bigl(y-1\bigr)^2-1=-1$$
$$\bigl(x-4\bigr)^2+\bigl(y-1\bigr)^2=16$$

Centre $$C_1=(4,1)$$, radius $$r_1=4$$.

Circle 2
$$x^2+y^2+6x+8y=0$$
Group and complete the squares:

$$\bigl(x^2+6x\bigr)+\bigl(y^2+8y\bigr)=0$$
$$\bigl(x+3\bigr)^2-9+\bigl(y+4\bigr)^2-16=0$$
$$\bigl(x+3\bigr)^2+\bigl(y+4\bigr)^2=25$$

Centre $$C_2=(-3,-4)$$, radius $$r_2=5$$.

Distance between centres
$$d=\sqrt{(4-(-3))^2+(1-(-4))^2}=\sqrt{7^2+5^2}=\sqrt{74}\approx8.602$$

Relative position of the circles
Sum of radii: $$r_1+r_2=4+5=9$$
Difference of radii: $$|r_1-r_2|=|4-5|=1$$

Since $$|r_1-r_2|\lt d\lt r_1+r_2$$, the circles intersect each other in two distinct points (they overlap).

Number of common tangents
For two intersecting circles, exactly two direct common tangents can be drawn; the transverse (cross) tangents do not exist because the circles overlap. Hence the total number of common tangents is $$2$$.

Option C which is: two