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If the circles x²+y²=2 and x²+y²–4x–4y+ $λ$ =0 have exactly three real common tangents the $λ$ =

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Detailed Solution

$C_{1} = (0, 0), r_{1} = \sqrt{2}, C_{2} = (2, 2), r_{2} = \sqrt{8 - λ} since the circles have exactlly three common tangents C_{1} C_{2} = r_{1} + r_{2} \Rightarrow \sqrt{8} = \sqrt{2} + \sqrt{8 - λ} \sqrt{8 - λ} = \sqrt{2} ∴ λ = 6$

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