$\text{ If }5x\left(1+\frac{1}{x^2+y^2}\right)=12\text{ and }5y\left(1-\frac{1}{x^2+y^2}\right)=4\forall x,y\in R,\text{ then correct option(s) }$is/are

$x=\frac{12}{15}\times \frac{1}{\left(1+\frac{1}{x^2+y^2}\right)}\mathrm{............}\left(i\right)$

$\text{ and }\mathrm{y}=\frac{4}{5}\times \left(\frac{1}{1-\frac{1}{{\mathrm{x}}^2+{\mathrm{y}}^2}}\right)(\dots \dots ...\text{ (ii) }$

$\text{ Put }{x}^2+y^2=\frac{1}{u}\text{ and add }\left(i{)}^2\text{ and }\right(\text{ ii }{)}^2$

$\Rightarrow \frac{1}{u}={\left(\frac{4}{5}\right)}^2\left(\frac{9}{{\left(1+u\right)}^2}+\frac{1}{{\left(1-u\right)}^2}\right)$

$\Rightarrow 25\left(u^2-2+\frac{1}{u^2}\right)=32\left\{5\left(u+\frac{1}{u}\right)-8\right\}$

$\Rightarrow 4-5or\frac{1}{5}$

$\Rightarrow \left(x,y\right)=\left(2,1\right)or\left(\frac{2}{5},\frac{-1}{5}\right)$