If the circles $x^2+y^2-4x+6y+8=0,x^2+y^2-10x-6y+14=0$ touch each other then the point of contact is

$C_1=(2,-3);C_2=(5,3)$

$C_1{C}_2=\sqrt{{(5-2)}^2+{(3+3)}^2}$

$=\sqrt{9+36}=\sqrt{45}$

$=3\sqrt{5}$

$r_1=\sqrt{4+9-8}=\sqrt{5}$

$r_2=\sqrt{25+9-14}=\sqrt{20}=2\sqrt{5}$

$r_1+r_2=3\sqrt{5}=C_1{C}_2$

The given circle touch externally

$\therefore$The point of contact divides $C_1{C}_2$ in the ratio $r_1:r_2=1:2$

$\therefore$The point of contact is $=(\frac{1\times 5+2\times 2}{1+2},\frac{1\times 3+2\times -3}{1+2})$

$=(3,-1)$