The condition for the circles x^2+y^2+ax+4=0 | vedclass

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(C) The given circles are $S_1: x^2+y^2+ax+4=0$ and $S_2: x^2+y^2+by+4=0$. The centers are $C_1 = (\frac{-a}{2}, 0)$ and $C_2 = (0, \frac{-b}{2})$. Th

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$\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{16}$

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(C) The given circles are $S_1: x^2+y^2+ax+4=0$ and $S_2: x^2+y^2+by+4=0$. The centers are $C_1 = (\frac{-a}{2}, 0)$ and $C_2 = (0, \frac{-b}{2})$. The radii are $r_1 = \sqrt{(\frac{a}{2})^2 - 4} = \frac{\sqrt{a^2-16}}{2}$ and $r_2 = \sqrt{(\frac{b}{2})^2 - 4} = \frac{\sqrt{b^2-16}}{2}$. For the circles to touch each other,the distance between centers $C_1C_2 = r_1 + r_2$. $\sqrt{(\frac{-a}{2}-0)^2 + (0 - (\frac{-b}{2}))^2} = \frac{\sqrt{a^2-16}}{2} + \frac{\sqrt{b^2-16}}{2}$. $\sqrt{\frac{a^2}{4} + \frac{b^2}{4}} = \frac{\sqrt{a^2-16} + \sqrt{b^2-16}}{2}$. $\sqrt{a^2+b^2} = \sqrt{a^2-16} + \sqrt{b^2-16}$. Squaring both sides: $a^2+b^2 = (a^2-16) + (b^2-16) + 2\sqrt{(a^2-16)(b^2-16)}$. $32 = 2\sqrt{(a^2-16)(b^2-16)} \Rightarrow 16 = \sqrt{(a^2-16)(b^2-16)}$. Squaring again: $256 = (a^2-16)(b^2-16) = a^2b^2 - 16a^2 - 16b^2 + 256$. $a^2b^2 = 16(a^2+b^2)$. Dividing by $16a^2b^2$: $\frac{1}{16} = \frac{a^2+b^2}{a^2b^2} = \frac{1}{b^2} + \frac{1}{a^2}$.

The condition for the circles $x^2+y^2+ax+4=0$ and $x^2+y^2+by+4=0$ to touch each other is

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