Question

Find the equation of circle circumcscribing the triangle whose sides are 3x – y – 9 = 0, 5x – 3y – 23 = 0 & x + y – 3 = 0.

Answer 1

L₁ L₂ + λL₂ L₃ + µL₁ L₃ = 0 

(3x – y – 9) (5x – 3y – 23) + λ(5x – 3y – 23) (x + y – 3) + µ (3x – y – 9) (x + y – 3) = 0 

(15x² + 3y² – 14xy – 114x + 50y + 207) + λ(5x² – 3y² + 2xy – 38x – 14y + 69) + µ (3x² – y² + 2xy – 18x – 6y + 27) = 0 

(5λ + 3µ + 15)x² + (3 – 3λ – µ)y² + xy (2λ + 2µ – 14) – x (114 + 38λ + 18µ) + y(50 – 14λ – 6µ) + (207 + 69λ + 27µ) = 0 ...........(i) 

coefficient of x² = coefficient of y² 

⇒ 5λ + 3µ + 15 = 3 – 3λ – µ 

8λ + 4µ + 12 = 0 

2λ + µ + 3 = 0 ...........(ii) 

coefficient of xy = 0 

⇒ 2λ + 2µ – 14 = 0 

⇒ λ + µ – 7 = 0 ..........(iii) 

Solving (ii) and (iii), we have 

λ = – 10, µ = 17 

Puting these values of λ & µ in equation (i), we get 

2x² + 2y² – 5x + 11y – 3 = 0