Equation of a circle with centre $(– 4, 3)$ touching internally and containing the circle $x^2+y^2=1$ is

Let the equation of the required circle be.
           $(x+4{)}^2+(y-3{)}^2=r^2$(i)

If (i) touches the circle $x^2+y^2=1$ internally         (ii)
the distance between the centres $(- 4, 3)$ and $(0, 0)$ of these circles is equal to the difference of their radii
$\Rightarrow \sqrt{4^2+3^2}=r-1\Rightarrow r=5+1\Rightarrow r=6$
So that an equation of the required circle is

$x^2+y^2+8x-6y-11=0.$