The set of all real values of lambda for whic | vedclass

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(B) The equations of the circles are:$C_1: x^2 + y^2 - 10x - 10y + \lambda = 0$ with center $O_1 = (5, 5)$ and radius $r_1 = \sqrt{5^2 + 5^2 - \lambda

Vedclass · Accepted Answer

$(18, 42)$

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(B) The equations of the circles are:$C_1: x^2 + y^2 - 10x - 10y + \lambda = 0$ with center $O_1 = (5, 5)$ and radius $r_1 = \sqrt{5^2 + 5^2 - \lambda} = \sqrt{50 - \lambda}$.$C_2: x^2 + y^2 - 4x - 4y + 6 = 0$ with center $O_2 = (2, 2)$ and radius $r_2 = \sqrt{2^2 + 2^2 - 6} = \sqrt{2}$.The distance between the centers is $d = O_1O_2 = \sqrt{(5-2)^2 + (5-2)^2} = \sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2}$.For exactly two common tangents,the circles must intersect at two distinct points,which implies $|r_1 - r_2| Substituting the values: $|\sqrt{50 - \lambda} - \sqrt{2}| This inequality splits into two parts:$1) \sqrt{50 - \lambda} - \sqrt{2}  18$.$2) \sqrt{50 - \lambda} + \sqrt{2} > 3\sqrt{2}$ $\Rightarrow \sqrt{50 - \lambda} > 2\sqrt{2}$ $\Rightarrow 50 - \lambda > 8$ $\Rightarrow \lambda Combining these,the interval for $\lambda$ is $(18, 42)$.

The set of all real values of $\lambda$ for which exactly two common tangents can be drawn to the circles $x^2 + y^2 - 4x - 4y + 6 = 0$ and $x^2 + y^2 - 10x - 10y + \lambda = 0$ is the interval:

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