In the given figure, the equation of the larger circle is x^ 2 +y^...

We have $x^{2}+y^{2}+4 y-5=0$. Its centre is $\mathrm{C}_{1}(0,-2)\mathrm{r}_{1}=\sqrt{4+5}=3$. Let $\mathrm{C}_{2}(\mathrm{ h}, \mathrm{k})$ be the centre of the smaller circle and its radius $\mathrm{r}_{2}$. Then $\mathrm{C}_{1} \mathrm{C}_{2}=4\Rightarrow \sqrt{\mathrmh^{2}+(\mathrm{k}+2)^{2}}=3+\mathrm{r}_{2}=4\Rightarrow \mathrm{r}_{2}=1$ But $\mathrm{k}=\mathrm{r}_{2}=1 \quad$ [it touches $\mathrm{x}$ -axis $\therefore$ From eq $(1), 4=\sqrt{\mathrmh^{2}+(1+2)^{2}}\Rightarrow 16=\mathrmh^{2}+9 \Rightarrow \mathrmh^{2}=7 \Rightarrow \mathrm{h}=\pm \sqrt{7}$ Since $\mathrm{h} > 0 \quad \therefore \mathrm{h}=\sqrt{7}$ Hence, required circle is $(x-\sqrt{7})^{2}+(y-1)^{2}=1$