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

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(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

Vedclass · Accepted Answer

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

Vedclass · Answer

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

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

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