If $\left({\mathrm{x}}_{\mathrm{i}},{\mathrm{y}}_{\mathrm{i}}\right), \mathrm{i}=1,2,3,4$ are concyclic points on $xy=4$ then ${\mathrm{x}}_1{\mathrm{x}}_2{\mathrm{x}}_3{\mathrm{x}}_4=$

(i)

Given hyperbola $\mathrm{xy}=4$

let $\mathrm{P}({\mathrm{x}}_{\mathrm{i}}, {\mathrm{y}}_{\mathrm{i}})=\left(2{\mathrm{t}}_{\mathrm{i}}, \frac{2}{{\mathrm{t}}_{\mathrm{i}}}\right)$ be any point on given hyperbola.

since given points are concyclic points then they lie on the circle ${\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{k}=0$

$$\begin{gathered} \Rightarrow 4{{\mathrm{t}}_{\mathrm{i}}}^2+\frac{4}{{{\mathrm{t}}_{\mathrm{i}}}^2}+2\mathrm{g}\left(2{\mathrm{t}}_{\mathrm{i}}\right)+2\mathrm{f}\left(\frac{2}{6_{\mathrm{i}}}\right)+\mathrm{k}=0 \\ \Rightarrow 4{{\mathrm{t}}_{\mathrm{i}}}^4+4{{\mathrm{gt}}_{\mathrm{i}}}^3+{{\mathrm{kt}}_{\mathrm{i}}}^2+4{\mathrm{ft}}_{\mathrm{i}}+4=0 \\ \mathrm{let} \mathrm{four} \mathrm{roots} \mathrm{be} {\mathrm{t}}_1, {\mathrm{t}}_2, {\mathrm{t}}_3, {\mathrm{t}}_4. \\ \mathrm{Now} {\mathrm{t}}_1 {\mathrm{t}}_2 {\mathrm{t}}_3 {\mathrm{t}}_4=\frac{4}{4}=1 \\ \Rightarrow {\mathrm{t}}_1 {\mathrm{t}}_2 {\mathrm{t}}_3 {\mathrm{t}}_4=1 \\ \mathrm{Now} \left({\mathrm{x}}_1 {\mathrm{x}}_2 {\mathrm{x}}_3 {\mathrm{x}}_4=\left(2{\mathrm{t}}_1\right) \left(2{\mathrm{t}}_2\right) \left(2{\mathrm{t}}_3\right) \left(2{\mathrm{t}}_4\right)\right) \\ =16\left({\mathrm{t}}_1{\mathrm{t}}_2{\mathrm{t}}_3{\mathrm{t}}_4\right) \\ =16.\left(1\right) \\ =16. \end{gathered}$$

(ii)

Given hyperbola $\mathrm{xy}={\mathrm{c}}^2$

let $\mathrm{P}({\mathrm{x}}_{\mathrm{i}}, {\mathrm{y}}_{\mathrm{i}})=\left({\mathrm{ct}}_{\mathrm{i}}, \frac{\mathrm{c}}{{\mathrm{t}}_{\mathrm{i}}}\right)$ where $\mathrm{i}=1, 2, 3, 4$ be point on given hyperbola.

since given points are concyclic points then they lie on the circle ${\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{k}=0$

$$\begin{gathered} \Rightarrow {\mathrm{c}}^2{{\mathrm{t}}_{\mathrm{i}}}^2+\frac{{\mathrm{c}}^2}{{{\mathrm{t}}_{\mathrm{i}}}^2}+2\mathrm{g}\left({\mathrm{t}}_{\mathrm{i}}\right)+2\mathrm{f}\left(\frac{\mathrm{c}}{{\mathrm{t}}_{\mathrm{i}}}+\mathrm{k}=0\right) \\ \Rightarrow {\mathrm{c}}^2{{\mathrm{t}}_{\mathrm{i}}}^4+2{{\mathrm{gct}}_{\mathrm{i}}}^3+{{\mathrm{kt}}_{\mathrm{i}}}^2+2{\mathrm{fct}}_{\mathrm{i}}+{\mathrm{c}}^2=0 \\ \mathrm{It} \mathrm{has} \mathrm{four} \mathrm{root}, \mathrm{let} \mathrm{it} \mathrm{be} {\mathrm{t}}_1, {\mathrm{t}}_2, {\mathrm{t}}_3, {\mathrm{t}}_4 \\ \mathrm{Now} {\mathrm{t}}_1{\mathrm{t}}_2{\mathrm{t}}_3{\mathrm{t}}_4=\frac{{\mathrm{c}}^2}{{\mathrm{c}}^2}=1 \\ \mathrm{Given} {\mathrm{y}}_1 {\mathrm{y}}_2 {\mathrm{y}}_3 {\mathrm{y}}_4=4 \\ \Rightarrow \left(\frac{\mathrm{C}}{{\mathrm{t}}_1}\right) \left(\frac{\mathrm{C}}{{\mathrm{t}}_2}\right) \left(\frac{\mathrm{C}}{{\mathrm{t}}_3}\right) \left(\frac{\mathrm{C}}{{\mathrm{t}}_4}\right)=4 \\ \Rightarrow \frac{{\mathrm{C}}^4}{{\mathrm{t}}_1 {\mathrm{t}}_2 {\mathrm{t}}_3 {\mathrm{t}}_4}=4 \\ \Rightarrow \frac{{\mathrm{C}}^4}{1}=4 \\ \Rightarrow {\mathrm{C}}^4=4 \\ \Rightarrow {\mathrm{c}}^2=2. \end{gathered}$$

(iii)

Given hyperbola $\mathrm{xy}={\mathrm{c}}^2$

$\Rightarrow \mathrm{x}=\frac{{\mathrm{c}}^2}{\mathrm{y}}.$

Given circle ${\mathrm{x}}^2+{\mathrm{y}}^2={\mathrm{r}}^2$

sub 'x' value

$$\begin{gathered} \Rightarrow \frac{{\mathrm{c}}^4}{{\mathrm{y}}^2}+{\mathrm{y}}^2={\mathrm{r}}^2 \\ \Rightarrow {\mathrm{c}}^4+{\mathrm{y}}^4={\mathrm{r}}^2{\mathrm{y}}^2 \\ \Rightarrow {\mathrm{y}}^4-{\mathrm{r}}^2{\mathrm{y}}^2+{\mathrm{c}}^4=0 \\ \mathrm{let} {\mathrm{y}}_1, {\mathrm{y}}_2, {\mathrm{y}}_3, {\mathrm{y}}_4 \mathrm{be} \mathrm{roots} \mathrm{of} \mathrm{this} \mathrm{equation} \mathrm{Now} {\mathrm{y}}_1+{\mathrm{y}}_2+{\mathrm{y}}_3+{\mathrm{y}}_4=0. \left(\therefore \mathrm{sum} \mathrm{of} \mathrm{roots}=\frac{-\mathrm{coefficient} \mathrm{of} {\mathrm{y}}^3}{\mathrm{coefficient} \mathrm{of} {\mathrm{y}}^4}\right) \end{gathered}$$