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(C) The given equations are $S_1: x^{2} + y^{2} - 2ax + c^{2} = 0$ and $S_2: x^{2} + y^{2} - 2by + c^{2} = 0$.For $S_1$,the center $C_1 = (a, 0)$ and

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

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(C) The given equations are $S_1: x^{2} + y^{2} - 2ax + c^{2} = 0$ and $S_2: x^{2} + y^{2} - 2by + c^{2} = 0$.For $S_1$,the center $C_1 = (a, 0)$ and radius $r_1 = \sqrt{a^{2} - c^{2}}$.For $S_2$,the center $C_2 = (0, b)$ and radius $r_2 = \sqrt{b^{2} - c^{2}}$.Two circles touch each other externally if the distance between their centers is equal to the sum of their radii,i.e.,$C_1C_2 = r_1 + r_2$.$C_1C_2 = \sqrt{(a-0)^{2} + (0-b)^{2}} = \sqrt{a^{2} + b^{2}}$.So,$\sqrt{a^{2} + b^{2}} = \sqrt{a^{2} - c^{2}} + \sqrt{b^{2} - c^{2}}$.Squaring both sides: $a^{2} + b^{2} = (a^{2} - c^{2}) + (b^{2} - c^{2}) + 2\sqrt{(a^{2} - c^{2})(b^{2} - c^{2})}$.$a^{2} + b^{2} = a^{2} + b^{2} - 2c^{2} + 2\sqrt{(a^{2} - c^{2})(b^{2} - c^{2})}$.$2c^{2} = 2\sqrt{(a^{2} - c^{2})(b^{2} - c^{2})}$.$c^{2} = \sqrt{(a^{2} - c^{2})(b^{2} - c^{2})}$.Squaring again: $c^{4} = (a^{2} - c^{2})(b^{2} - c^{2}) = a^{2}b^{2} - a^{2}c^{2} - b^{2}c^{2} + c^{4}$.$a^{2}c^{2} + b^{2}c^{2} = a^{2}b^{2}$.Dividing by $a^{2}b^{2}c^{2}$,we get $\frac{1}{b^{2}} + \frac{1}{a^{2}} = \frac{1}{c^{2}}$.

When do the circles represented by the equations $x^{2} + y^{2} + c^{2} = 2ax$ and $x^{2} + y^{2} + c^{2} = 2by$ touch each other externally?

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