Mathematics - Circle Question with Solution | TestHub

Since $|\mathrm{r}|>1,\frac{1}{|\mathrm{r}|}<1$ $\therefore \mathrm{x}=\frac{\mathrm{a}}{1-\frac{1}{\mathrm{r}}}=\frac{\mathrm{ar}}{\mathrm{r}-1}$ Similarly, $y=\frac{b}{1-\left(-\frac{1}{r}\right)}=\frac{br}{r+1}$ and $\begin{array}{c}z=\frac{c}{1-\frac{1}{r^2}}=\frac{c{r}^2}{r^2-1}...(1)\\ \therefore xy=\frac{ar}{r-1}\times \frac{br}{r+1}=\frac{ab{r}^2}{r^2-1}...(2)\end{array}$ Dividing (2) by (1), we get $\frac{\mathrm{xy}}{\mathrm{z}}=\frac{{\mathrm{abr}}^2}{{\mathrm{r}}^2-1}\times \frac{{\mathrm{r}}^2-1}{{\mathrm{cr}}^2}=\frac{\mathrm{ab}}{\mathrm{c}}$