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.

Question

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.

Sher Mohammad · Answer

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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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.

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