The point/points of intersection of common tangents of the two circles $x^2+y^2-8x-6y+21=0$ and $x^2+y^2-2y-15=0$ is/are

$x^2+y^2-8x-6y+21=0\stackrel{ }{\to }\left(1\right)$

$c_1=\left(4,3\right);r_1=\sqrt{16+9-21}=2$

$x^2+y^2-2y-15=0\stackrel{ }{\to }\left(2\right)$

$c_2=\left(0,1\right);r_2=\sqrt{0+1+15}=\sqrt{16}=4$

$d=c_1{c}_2=\sqrt{16+4}=\sqrt{20}$

$r_1+r_2=2+4=6=\sqrt{36}$

$\therefore c_1{c}_2<r_1+r_2$

$\therefore$ (1), (2) intersect in two distinct points.
$\therefore$ no. of common tangents = 2

$\therefore$ P.I of common tangent is External centre of similitude 
$r_1:r_2=2:4=1:2$
E.C.S.= $\frac{r_1{c}_2-r_2{c}_1}{r_1-r_2}=\frac{1\left(0,1\right)-2\left(4,3\right)}{1-2}$
$=\left(\frac{-8}{-1},\frac{-5}{-1}\right)=\left(8,5\right)$