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

Refer The Attachment ⬆️

$\huge\underbrace{\pink{\sf{SoluTion:}}}$

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' 

Hence, Proved f'g = fg'✔️

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$\red{\sf{\star\; MrMonarque}}$

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