Помогите решить пожалуйста

Question 1

Question 2

$\left \{ {{ \sqrt{x^2+8}+ \sqrt{y^2-8} =6} \atop {x^2+y^2=20}} \right.;\\ \left \{ {{ \sqrt{x^2+8}+ \sqrt{20-x^2-8} =6} \atop {y^2=20-x^2}} \right.;\\ \left \{ {{ \sqrt{x^2+8}+ \sqrt{12-x^2} =6\ \ \ (1)} \atop {y^2=20-x^2\ \ \ (2)}} \right.;\\$

Решим отдельно уравнение (1):
$\sqrt{x^2+8}+ \sqrt{12-x^2} =6\\ \left \{ {{\sqrt{t+8}+ \sqrt{12-t} =6} \atop {x^2=t \geq 0}} \right.\\ \left \{ {({\sqrt{t+8}+ \sqrt{12-t})^2 =6^2} \atop {x^2=t \geq 0\ and \ t+8 \geq 0\ and\ 12-t \geq 0}} \right.\\ \left \{ {({\sqrt{t+8})^2+ 2*\sqrt{t+8}*\sqrt{12-t}+(\sqrt{12-t})^2 =36} \atop {x^2=t \geq 0\ and \ t \geq -8\ and\ t \leq 12}} \right.\\ \left \{ {t+8+ 2\sqrt{(t+8)(12-t)}+12-t =36} \atop {x^2=t\ and \ 0\leq t \leq 12}} \right.\\$

$\left \{ {\sqrt{(t+8)(12-t)}=8} \atop {x^2=t\ and \ 0\leq t \leq 12}} \right.\\ \left \{ (t+8)(12-t)=64} \atop {x^2=t\ and \ 0\leq t \leq 12}} \right.\\ \left \{ -t^2+4t+96=64} \atop {x^2=t\ and \ 0\leq t \leq 12}} \right.\\ \left \{ t^2-4t-32=0} \atop {x^2=t\ and \ 0\leq t \leq 12}} \right.\\ \left \{ t^2-4t-32=0} \atop {x^2=t\ and \ 0\leq t \leq 12}} \right.\\ \left \{ t=8\ or\ t=-4} \atop {x^2=t\ and \ 0\leq t \leq 12}} \right.\\$

$x^2=8;\\ x=2 \sqrt{2}\ or\ x=-2 \sqrt{2}$

Возвращаемся к начальной системе и имеем:
$\left \{ {{y^2=20-x^2=20-8=12} \atop {x=2 \sqrt{2}\ or\ x=-2 \sqrt{2}}} \right.;\\ \left \{ {{y=2 \sqrt{3}\ or\ y=-2 \sqrt{3}}} \atop {x=2 \sqrt{2}\ or\ x=-2 \sqrt{2}}} \right.;$

Ответ: $(2 \sqrt{2};\ 2 \sqrt{3} ),\ (2 \sqrt{2};\ -2 \sqrt{3} ); (-2 \sqrt{2};\ 2 \sqrt{3} ); (-2 \sqrt{2};\ -2 \sqrt{3} )$