If the circles x^2 + y^2 = 4 and x^2 + y^2 - | vedclass

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(C) The given circles are $x^2 + y^2 = 4$ and $x^2 + y^2 - 10x + \lambda = 0$.For the first circle,the center $C_1 = (0, 0)$ and radius $r_1 = \sqrt{4

Vedclass · Accepted Answer

$16$

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(C) The given circles are $x^2 + y^2 = 4$ and $x^2 + y^2 - 10x + \lambda = 0$.For the first circle,the center $C_1 = (0, 0)$ and radius $r_1 = \sqrt{4} = 2$.For the second circle,the center $C_2 = (5, 0)$ and radius $r_2 = \sqrt{5^2 - \lambda} = \sqrt{25 - \lambda}$.Since the circles touch externally,the distance between their centers must be equal to the sum of their radii: $C_1C_2 = r_1 + r_2$.The distance $C_1C_2 = \sqrt{(5-0)^2 + (0-0)^2} = 5$.So,$5 = 2 + \sqrt{25 - \lambda}$.$3 = \sqrt{25 - \lambda}$.Squaring both sides,$9 = 25 - \lambda$.Therefore,$\lambda = 25 - 9 = 16$.

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

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