$\{\begin{array}{l}{x}^2+y^2=5\\ y=-2x-3\end{array}$  

$\{\begin{array}{l}{x}^2+{\left(-2x-3\right)}^2=5\\ y=-2x-3\end{array}$                              

$\{\begin{array}{l}{x}^2+{\left(-\left(2x+3\right)\right)}^2=5\\ y=-2x-3\end{array}$  

$\{\begin{array}{l}{x}^2+4x^2+12x+9=5\\ y=-2x-3\end{array}$  

$\{\begin{array}{l}5x^2+12x+4=0\\ y=-2x-3\end{array}$  

$\Delta ={12}^2-4\cdot 5\cdot 4=144-80=64$  

$x_1=\frac{-12-8}{2\cdot 5}=\frac{-20}{10}=-2$  

$x_2=\frac{-12+8}{2\cdot 5}=\frac{-4}{10}=-\frac{2}{5}$  

Zatem mamy:

$\{\begin{array}{l}x=-2\\ y=-2x-3\end{array}\vee \{\begin{array}{l}x=-\frac{2}{5}\\ y=-2x-3\end{array}$  

$\{\begin{array}{l}x=-2\\ y=-2\cdot \left(-2\right)-3\end{array}\vee \{\begin{array}{l}x=-\frac{2}{5}\\ y=-2\cdot \left(-\frac{2}{5}\right)-3\end{array}$  

$\{\begin{array}{l}x=-2\\ y=4-3\end{array}\vee \{\begin{array}{l}x=-\frac{2}{5}\\ y=\frac{4}{5}-3\end{array}$  

$\{\begin{array}{l}x=-2\\ y=1\end{array}\vee \{\begin{array}{l}x=-\frac{2}{5}\\ y=-2\frac{1}{5}\end{array}$  

$A=\left(-2,1\right),B=\left(-\frac{2}{5},-2\frac{1}{5}\right)$  

 

Interpretacja geometryczna:

Wystarczy narysować w jednym układzie współrzędnych:

1) Okrąg o środku S(0, 0) i promieniu r=√5.

2) Prostą y=-2x-3.