Intersections 3

Find the intersections of the circles x² + y² + 6 x - 10 y + 9 = 0 and x² + y² + 18 x + 4 y + 21 = 0

Correct answer:

x₁ = -1.3335
y₁ = 0.2859
x₂ = -7.9135
y₂ = 5.9259

x^{2} + y^{2} + 6 x - 10 y + 9 = 0 x^{2} + y^{2} + 18 x + 4 y + 21 = 0 e_{2} - e_{1} : (18 - 6) x + (4 + 10) y + (21 - 9) = 0 y = - 6 / 7 (x + 1) x^{2} + (- 6 / 7 (x + 1))^{2} + 6 x - 10 (- 6 / 7 (x + 1)) + 9 = 0 x^{2} + (- 6 / 7 (x + 1))^{2} + 6 x - 10 (- 6 / 7 (x + 1)) + 9 = 0 1.734693877551 x^{2} + 16.041 x + 18.306 = 0 a = 1.734694; b = 16.041; c = 18.306 D = b^{2} - 4 a c = 16.04 1^{2} - 4 \cdot 1.734694 \cdot 18.306 = 130.2857142857 D > 0 x_{1, 2} = \frac{- b \pm D}{2 a} = \frac{- 1 6 . 0 4 \pm 1 3 0 . 2 9}{3 . 4 6 9 3 8 8} x_{1, 2} = - 4.623529 \pm 3.289997 x_{1} = - 1.333531989 = - 1.3335 x_{2} = - 7.913526835

y_{1} = - 6 / 7 \cdot (x_{1} + 1) = - 6 / 7 \cdot ((- 1.3335) + 1) = 0.2859

y_{2} = - 6 / 7 \cdot (x_{2} + 1) = - 6 / 7 \cdot ((- 7.9135) + 1) = 5.9259

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