JEE Advanced 2022 — Paper 1 — Question 8

Let A be $(0,0), \mathrm{B}(1,0)$ and $\mathrm{C}(0,3)$

$\therefore \mathrm{AB}$ lies on x -axis and AC lies on y -axis

$\therefore$ equation of circle touching both x and y -axis is of the form

$(\mathrm{x}-\mathrm{h})^{2}+(\mathrm{y}-\mathrm{h})^{2}=\mathrm{h}^{2} \quad(\because \mathrm{~h}=\mathrm{k}=\mathrm{r})$

It touches the circle $\left(x-\frac{1}{2}\right)^{2}+\left(y-\frac{3}{2}\right)^{2}=\frac{5}{2}$

$\therefore \mathrm{c}_{1} \mathrm{c}_{2}=\left|\mathrm{r}_{1}-\mathrm{r}_{2}\right|$

$\sqrt{\left(\mathrm{h}-\frac{1}{2}\right)^{2}+\left(\mathrm{h}-\frac{3}{2}\right)^{2}}=\left|\mathrm{h}-\frac{\sqrt{5}}{\sqrt{2}}\right|$

$\Rightarrow \mathrm{h}^{2}+\frac{1}{4}-\mathrm{h}+\mathrm{h}^{2}+\frac{9}{4}-3 \mathrm{~h}=\mathrm{h}^{2}+\frac{5}{2}-\sqrt{10} \mathrm{~h}$

$\Rightarrow \mathrm{h}^{2}+(\sqrt{10}-4) \mathrm{h}=0 \Rightarrow \mathrm{~h}=4-\sqrt{10}$

$\therefore \mathrm{r}=4-\sqrt{10}=0.8377$