Answer: A = (-2, 0) B = (-1, 1) x² + y² = 1/2
Explanation:
To find where the line and circle intersect, create a system of equations and use the Substitution method to find the coordinates:
EQ1: x - y + 2 = 0 → x = y - 2
EQ2: x² + 2x + y² = 0 → (y - 2)² + 2(y - 2) + y² = 0
→ (y² - 4y + 4) + (2y - 4) + y² = 0
→ 2y² - 2y = 0
→ 2y(y - 1) = 0
→ 2y = 0 y - 1 = 0
y = 0 y = 1
Plug the y-values into EQ1 to solve for x:
x = y - 2 x = y - 2
x = 0 - 2 x = 1 - 2
x = - 2 x = - 1
Coordinates: (-2, 0) & (-1, 1)
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Diameter of AB: d² = (x₂ - x₁)² + (y₂ - y₁)²
= [-2 - (-1)]² + [0 - (-1)]²
= (-2 + 1)² + (0 + 1)²
= (-1)² + 1²
= 1 + 1
d² = 2
d = √2
→ r = √2/2
Equation of a circle is: (x - h)² + (y - k)² = r²
Since a center was not provided for the equation with diameter AB, you can choose any value for center (h, k). I choose to let (h, k) = (0, 0) (x - 0)² + (y - 0)² = (√2/2)²
x² + y² = 2/4
x² + y² = 1/2