Given that, equation of circle
$x^{2}+y^{2}=1$
Centre at $O \rightarrow(0,0)$
Radius $=O A=1$
Also, the centre of another circle $\rightarrow C(4,3)$ both circle touch externally. Then, distance between centres $=O C$.
$=\sqrt{(4-0)^{2}+(3-0)^{2}}=\sqrt{16-9}=5$
Now, $A C=O C-O A$
$A C=5-1=4$
So, the radius of other circle is 4 .
Now, the equation of other circle touch externally to the circle $x^{2}+y^{2}=1$ is,
$(x-4)^{2}+(y-3)^{2}=16$
$\Rightarrow x^{2}+y^{2}-8 x-6 y+9=0$