The number of common tangent to the circle $x^2+y^2=9$ and $x^2+y^2-6 x-8 y+21=0$

If equation of circle is $x^2+y^2+2 g x+2 f y+c=0$
Then, centre $\equiv(-g,-f)$
$\text { radius }=\sqrt{g^2+f^2-c}$
$\Rightarrow$ Centre of first circle $\equiv(0,0)$
radius of first circle $\equiv 3$
$\Rightarrow$ Centre of second circle $\equiv(3,4)$
radius of second circle $\equiv \sqrt{9+16-21}=2$
$\because$ Distance between centres of circle $\left(c_1 c_2\right)$
$=\sqrt{3^2+4^2}=5$
$\therefore$ Sum of radius $\left(r_1+r_2\right)=5$
$c_1 c_2=r_1+r_2$ (circle touches externally)
$\therefore$ Number of common tangent $=3$