if the line 2x + 4y + 8 = 0 is a diameter of a circle x2 + y2 – 2x + 6y = 0 then k= About the author Amelia
Answer: pls mark me as brainliest if you are satisfied Step-by-step explanation: Required equation of circle passing through intersection S = 0 and L = 0 is S + λL = 0 (x2 + y2 – 2x + 4y – 8) + λ(x + y – 3) = 0 (x2 + y2 + x(–2 + λ) + y (4 + λ) – 8 – 3λ = 0 —— (i) x2 + y2 + 2gx + 2fy + c = 0 —— (ii) Comparing (i) and (ii) we get g = (-2 + λ)/2, f = (4 + λ)/2 Centre lies on x + y = 3 ∴ – ((-2 + λ)/2) – ((4 + λ)/2) = 3 2 – λ – 4 – λ = 6 –2λ = 8 ⇒ λ = – 4 Required equation of circle be (x2 + y2 – 2x + 4y – 8) – 4(x + y – 3) = 0 x2 + y2 – 6x + 4 = 0 Reply
Answer:
pls mark me as brainliest if you are satisfied
Step-by-step explanation:
Required equation of circle passing through intersection S = 0 and L = 0 is S + λL = 0 (x2 + y2 – 2x + 4y – 8) + λ(x + y – 3) = 0 (x2 + y2 + x(–2 + λ) + y (4 + λ) – 8 – 3λ = 0 —— (i) x2 + y2 + 2gx + 2fy + c = 0 —— (ii) Comparing (i) and (ii) we get g = (-2 + λ)/2, f = (4 + λ)/2 Centre lies on x + y = 3 ∴ – ((-2 + λ)/2) – ((4 + λ)/2) = 3 2 – λ – 4 – λ = 6 –2λ = 8 ⇒ λ = – 4 Required equation of circle be (x2 + y2 – 2x + 4y – 8) – 4(x + y – 3) = 0 x2 + y2 – 6x + 4 = 0