If the circles x^2 + y^2 + 2ax + c = 0 and x^ | vedclass

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(D) The centers of the circles are $C_1(-a, 0)$ and $C_2(0, -b)$.The radii of the circles are $R_1 = \sqrt{a^2 - c}$ and $R_2 = \sqrt{b^2 - c}$.The di

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

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(D) The centers of the circles are $C_1(-a, 0)$ and $C_2(0, -b)$.The radii of the circles are $R_1 = \sqrt{a^2 - c}$ and $R_2 = \sqrt{b^2 - c}$.The distance between the centers is $C_1C_2 = \sqrt{(-a - 0)^2 + (0 - (-b))^2} = \sqrt{a^2 + b^2}$.Since the circles touch each other,the distance between their centers must be equal to the sum of their radii: $R_1 + R_2 = C_1C_2$.$\sqrt{a^2 - c} + \sqrt{b^2 - c} = \sqrt{a^2 + b^2}$.Squaring both sides: $(a^2 - c) + (b^2 - c) + 2\sqrt{(a^2 - c)(b^2 - c)} = a^2 + b^2$.$a^2 + b^2 - 2c + 2\sqrt{(a^2 - c)(b^2 - c)} = a^2 + b^2$.$2\sqrt{(a^2 - c)(b^2 - c)} = 2c$.$\sqrt{(a^2 - c)(b^2 - c)} = c$.Squaring again: $(a^2 - c)(b^2 - c) = c^2$.$a^2b^2 - a^2c - b^2c + c^2 = c^2$.$a^2b^2 = a^2c + b^2c$.Dividing both sides by $a^2b^2c$: $\frac{a^2b^2}{a^2b^2c} = \frac{a^2c}{a^2b^2c} + \frac{b^2c}{a^2b^2c}$.$\frac{1}{c} = \frac{1}{b^2} + \frac{1}{a^2}$.

If the circles $x^2 + y^2 + 2ax + c = 0$ and $x^2 + y^2 + 2by + c = 0$ touch each other,then:

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