If the circles x^{2}+y^{2}+6 x+8 y+16=0 and x^{2}+y^{2}+2(3-\sqrt{3}) x+2(4-\sqrt{6}) y=k+6 \sqrt{3}+8 \sqrt{6}, k>0, touch internally at the point P(\alpha, \beta), then (\alpha+\sqrt{3})^{2}+(\beta+\sqrt{6})^{2} is equal to [25-Jul-2022-Shift-2]

JEE Mains 25-July-2022 Shift 2 Solved Paper

Section: Mathematics

Question : 28 of 90

Marks: +1, -0

If the circles x2+y2+6x+8y+16=0 and x2+y2+2(3−√3)x+2(4−√6)y=k+6√3+8√6,k>0, touch internally at the point P(α,β), then (α+√3)2+(β+√6)2 is equal to

[25-Jul-2022-Shift-2]

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