Equation of a circle with centre $(4, 3)$ touching the circle

$x^2+y^2=1$, is

Let the circle be $x^2+y^2-8x-6y+k=0$ which touches the circle $x^2+y^2=1.$ The equation of the common tangent to these circles is given by  

$S_1-S_2=0\Rightarrow 8x+6y-1-k=0$

This is a tangent to the circle $x^2+y^2=1$

$\therefore \pm 1=\frac{k+1}{\sqrt{8^2+6^2}}\Rightarrow k^2+1=\pm 10\Rightarrow k=-11$ or, $9$

Hence, the equations of the circles are

$x^2+y^2-8x-6y+9=0$and, $x^2+y^2-8x-6y-11=0$