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JEE MATHEMATICS MCQ: CIRCLES

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Application +4 | -1
If the common chord of the circles x2+(y−b)2=16x^2+(y-b)^2=16 and x2+y2=16x^2+y^2=16 subtends a right angle at the origin, then b=b=
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Correct option: C
±42\pm 4 \sqrt{2}
Option C (±42\pm 4 \sqrt{2}) is correct because it is the value of b for which the common chord subtends a right angle at the origin.

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