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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.
x^2 + y^2 + 2(a1)x +a1^2-a1^2+b = 0center 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-ba2^2>a1^2 >0 imply a2>a1>0 hence a1a2>0and both radius a2^2>=b>0,a1^2>=b>0 imply b>0a1^2-2a1a2+b<0 sher mohammadb.tech, iit delhi
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