Question

If the two circles x² + y² + 2gx + 2fy = 0 and x² + y² + 2g'x + 2f'y = 0 touch each other then show that f'g = fg'.

Answer 1

C₁ = (–g, –f) 

C₂ = (–g^–1, f^–1) 

r₁ = √(g² + f²) 

r₂ = √(g'² + r'²) 

C₁ C₂ = r₁ + r₂ 

(C₁ C₂)² = (r₁ + r₂)² 

(g' – g)² + (f' – f)² = g² + f² + g'² + f'² + 2√(g² + f²) √(g² + r²)

–2(gg' + ff') = 2{g² g'² + f² f'² + g² f'² + f² g'² }^1/2 

Squaring again 

(gg' + ff')² = g² g'² + f² f'² + g² f'² + g'² f² 

g² g'² + f² f'² + 2gg'ff' = g² g'² + f² f'² + g² f'² + g'² f'² 

2gg'ff' = g² f'² + f² g'² 

⇒ g² f'² + g'² f² – 2gg'ff' = 0 

(or) (gf' – fg')² = 0 

(or) gf' = fg'