In the given figure, the equation of the larger circle is $x^{2}+y^{2}+4 y-5=0$ and the distance between…
In the given figure, the equation of the larger circle is $x^{2}+y^{2}+4 y-5=0$ and the distance between centres is 4 . Then the equation of smaller circle is
$(x-\sqrt{7})^{2}+(y-1)^{2}=1$
$\quad(x+\sqrt{7})^{2}+(y-1)^{2}=1$
$x^{2}+y^{2}=2 \sqrt{7} x+2 y$
None of these
Solution
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 $\mathrm{C}_{1} \mathrm{C}_{2}=4$ $\Rightarrow \sqrt{\mathrm{h}^{2}+(\mathrm{k}+2)^{2}}=3+\mathrm{r}_{2}=4$
$\Rightarrow \mathrm{r}_{2}=1$
But $\mathrm{k}=\mathrm{r}_{2}=1 \quad$ [it touches $\mathrm{x}$ -axis
$\therefore$ From eq $(1), 4=\sqrt{\mathrm{h}^{2}+(1+2)^{2}}$
$\Rightarrow 16=\mathrm{h}^{2}+9 \Rightarrow \mathrm{h}^{2}=7 \Rightarrow \mathrm{h}=\pm \sqrt{7}$
Since $\mathrm{h}>0 \quad \therefore \mathrm{h}=\sqrt{7}$
Hence, required circle is $(x-\sqrt{7})^{2}+(y-1)^{2}=1$
