In the given figure, the equation of the large circle is $x^2+y^2+4y-5=0$  and the distance between centres is 4. Then the equation of smaller circle is::

Since $x^2+y^2+4y-5=0$

Centre is $C_1(0,-2)$ and Radius $r_1=\sqrt{4+5}=3$

Let $C_2(h,k)$  be the centre of the smaller circle and its radius $r_2$.

Since, $C_1{C}_2=4$

$\Rightarrow \sqrt{h^2+{(k+2)}^2}=3+r_2=4\Rightarrow r_2=1$

But $k=r_2=1$

Hence, using equation (1), we have     $\therefore h=\pm \sqrt{7}$

Since, h > 0

Hence, required circle is  ${(x-\sqrt{7})}^2+{(y-1)}^2=2$