Mathematics - Circle Question with Solution | TestHub

We have $x^2+y^2+4y-5=0$. 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 $\mathrm{k}={\mathrm{r}}_2=1$ [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 $\mathrm{h}>0\therefore \mathrm{h}=\sqrt{7}$ Hence, required circle is $(x-\sqrt{7}{)}^2+(y-1{)}^2=1$