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If a circle x^2 + y^2 + 2(a1)x + b = 0 lies completely inside the circle x^2 + y^2 + 2(a2)x + b = 0, then prove that (a1)(a2) > 0 & b > 0.

If a circle x^2 + y^2 + 2(a1)x + b = 0 lies completely inside the circle x^2 + y^2 + 2(a2)x + b = 0, then prove that (a1)(a2) > 0 & b > 0.

Grade:12th pass

1 Answers

Sher Mohammad IIT Delhi
askIITians Faculty 174 Points
8 years ago
x^2 + y^2 + 2(a1)x +a1^2-a1^2+b = 0
center of first circle is (-a1,0) and radius is sqrt(a1^2-b)
a1^2-2a1a2+b<0 (centre of first circle lies inside second circle)

a2^2-b>a1^2-b
a2^2>a1^2 >0 imply a2>a1>0 hence a1a2>0
and both radius a2^2>=b>0,a1^2>=b>0 imply b>0
a1^2-2a1a2+b<0


sher mohammad
b.tech, iit delhi


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