If one of the diameters of the circle x^2+y^2-2 \sqrt{2} x-6 \sqrt{2} y+14=0 is a chord of the circle (x-2 \sqrt{2})^2+(y-2 \sqrt{2})^2=r^2, then the value of r^2 is equal to_____ [28-Jun-2022-Shift-2]

JEE Main 28 June 2022 Shift 2 Solved Paper

Section: Mathematics

Question : 23 of 90

Marks: +1, -0

If one of the diameters of the circle x2+y2−2√2x−6√2y+14=0 is a chord of the circle (x−2√2)2+(y−2√2)2=r2, then the value of r2 is equal to_____

[28-Jun-2022-Shift-2]

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