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If circles ${x^2} + {y^2} + 2ax + c = 0$and ${x^2} + {y^2} + 2by + c = 0$ touch each other, then
$\frac{1}{a} + \frac{1}{b} = \frac{1}{c}$
$\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} = \frac{1}{{{c^2}}}$
$\frac{1}{a} + \frac{1}{b} = {c^2}$
$\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} = \frac{1}{c}$
Solution
(d) ${C_1}( – a,\;0);$${C_2}(0,\; – b);$
${R_1}(\sqrt {{a^2} – c} );$ ${R_2}(\sqrt {{b^2} – c} )$
${C_1}{C_2} = \sqrt {{a^2} + {b^2}} $
Since they touch each other, tiply by $\frac{1}{{{a^2}{b^2}c}},\;$
we get $\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} = \frac{1}{c}$.
therefore $\sqrt {{a^2} – c} + \sqrt {{b^2} – c} = \sqrt {{a^2} + {b^2}} $
$ \Rightarrow {a^2}{b^2} – {b^2}c – {a^2}c$= 0
Multiply by $\frac{1}{{{a^2}{b^2}c}},\;$
we get $\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} = \frac{1}{c}$
