Question

Dadas as circunferências λ1 : x2 + y2 -2x -3 = 0 e λ2 : x2 + y2 + 2x - 4y + 1 = 0, determine seus pontos de interseção.

Answer 1

$\displaystyle \lambda_1 : \text x^2+\text y^2-2\text x-3=0 \\\\ \lambda_2: \text x^2+\text y^2+2\text x-4\text y+1=0 \\\\ \underline{\text{Pontos de interse{\c c}{\~a}o}}:\\\\ \text x^2+\text y^2-2\text x+3 = \text x^2+\text y^2+2\text x-4\text y+1\\\\ -2\text x-3=2\text x-4\text y+1 \\\\ -4\text x=-4\text y +4 \\\\ \boxed{\text x = \text y-1} \\\\ \underline{\text{Substituindo esse valor em }\lambda_1}: \\\\ (\text y-1)^2+\text y^2-2(\text y-1)-3=0 \\\\ \text y^2-2\text y+1+\text y^2-2\text y+2-3=0$

$2\text y^2-4\text y=0 \\\\ 2\text y(\text y-2)=0 \\\\ \boxed{\text y = 0} \ , \boxed{ \text y = 2}$

Daí :

$\text x =\text y-1 \\\\ \text x = 0-1 \to \boxed{\text x=-1} \\\\ \text x = 2-1 \to \boxed{\text x =1}$

Portanto os pontos de interseção são :

$\huge\boxed{\text x = -1 \ ; \ \text y = 0\ }\checkmark\\\\\\\ \huge\boxed{\text x = 1 \ ; \ \text y = 2\ }\checkmark$