If the circles {x^2} + {y^2} - 9 = 0 and {x^2 | vedclass

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Question

(a) When two circles touch each other externally, then

${r_1} + {r_2} = \sqrt {{{\left\{ {0 - ( - a)} \right\}}^2} + {{\left\{ {0 - ( - 1)} \right\

Vedclass · Accepted Answer

$(a)$ and $(b)$ both

Vedclass · Answer

(a) When two circles touch each other externally, then

${r_1} + {r_2} = \sqrt {{{\left\{ {0 - ( - a)} \right\}}^2} + {{\left\{ {0 - ( - 1)} \right\}}^2}} $

$ \Rightarrow (3 + a) = \sqrt {{a^2} + 1}$

$\Rightarrow a = - 4/3$.

When two circles touch each other internally, then

${r_1} - {r_2} = \sqrt {{{\left\{ {0 - ( - a)} \right\}}^2} + {{\left\{ {0 - ( - 1)} \right\}}^2}} $

$ \Rightarrow (3 - a) = \sqrt {{a^2} + 1}$

$\Rightarrow a = 4/3$.

If the circles ${x^2} + {y^2} - 9 = 0$ and ${x^2} + {y^2} + 2ax + 2y + 1 = 0$ touch each other, then $a =$

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