If circles {x^2} + {y^2} = 4,{x^2} + {y^2} - 10x + λ = 0 touch externally, then λ is equal to&nbsp

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10-1.Circle and System of Circles

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If the circles ${x^2} + {y^2} = 4,{x^2} + {y^2} - 10x + \lambda = 0$ touch externally, then $\lambda $ is equal to

A

$-16$

B

$9$

C

$16$

D

$25$

Solution

(a) Circles $x^2$ + $y^2$ = 4, $x^2$ + $y^2$ -10x + $\lambda$ = $0$ touch externally

${C_1}{C_2} = {r_1} + {r_2}$

==> ${C_1}(0,\,0)$ and ${C_2} = (5,\,0)$

${r_1} = 2$ and ${r_2} = \sqrt {25 + \lambda } $

$\sqrt {{{(5 – 0)}^2} + 0} = 2 + \sqrt {25 + \lambda } $

==> $5 – 2 = \sqrt {25 + \lambda } $

==> $3 = \sqrt {25 + \lambda } $

==> $9 = 25 + \lambda $

==> $\lambda$ $= -16.$

Standard 11

Mathematics

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