If one of the two circles x^2+y^2+alpha_1(x-y | vedclass

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(C) The equations of the given circles are $x^2+y^2+\alpha_1(x-y)+c=0$ and $x^2+y^2+\alpha_2(x-y)+c=0$.The centers are $C_1 = (-\frac{\alpha_1}{2}, \f

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$c > 0$

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(C) The equations of the given circles are $x^2+y^2+\alpha_1(x-y)+c=0$ and $x^2+y^2+\alpha_2(x-y)+c=0$.The centers are $C_1 = (-\frac{\alpha_1}{2}, \frac{\alpha_1}{2})$ and $C_2 = (-\frac{\alpha_2}{2}, \frac{\alpha_2}{2})$.The radii are $r_1 = \sqrt{\frac{\alpha_1^2}{4} + \frac{\alpha_1^2}{4} - c} = \sqrt{\frac{\alpha_1^2}{2} - c}$ and $r_2 = \sqrt{\frac{\alpha_2^2}{2} - c}$.For one circle to lie within the other,the distance between centers must be less than the difference of the radii: $d(C_1, C_2) The distance $d = \sqrt{(-\frac{\alpha_1}{2} + \frac{\alpha_2}{2})^2 + (\frac{\alpha_1}{2} - \frac{\alpha_2}{2})^2} = \sqrt{2(\frac{\alpha_1-\alpha_2}{2})^2} = \frac{|\alpha_1-\alpha_2|}{\sqrt{2}}$.Squaring both sides: $d^2 $\frac{(\alpha_1-\alpha_2)^2}{2} $\frac{\alpha_1^2 + \alpha_2^2 - 2\alpha_1\alpha_2}{2} $-\alpha_1\alpha_2 Squaring again: $4(\frac{\alpha_1^2}{2}-c)(\frac{\alpha_2^2}{2}-c) $(\alpha_1^2-2c)(\alpha_2^2-2c) $\alpha_1^2\alpha_2^2 - 2c(\alpha_1^2+\alpha_2^2) + 4c^2 $-2c(\alpha_1^2+\alpha_2^2+2\alpha_1\alpha_2) Wait,re-evaluating the condition $d  0$ given $\alpha_1 
eq \alpha_2$.

If one of the two circles $x^2+y^2+\alpha_1(x-y)+c=0$ and $x^2+y^2+\alpha_2(x-y)+c=0$ lies within the other,then (where $\alpha_1, \alpha_2 \in R, \alpha_1 \neq \alpha_2$):

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