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Question

If the circle x2+y2+2x+2ky+6=0 and x2+y2+2ky+k=0 intersect orthogonality, then k is:

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Solution

We have,
x2+y2+2x+2ky+6=0x2+y2+2ky+k=02×1×0+2×k×k=6+k2k2=k+62k2=k+62k2k6=02k24k+3k6=0(2k+3)(k2)=0k=32k=2

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