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Question

The value of k for which the centres of the circles x2 + y2 = 1, x2 + y2 + 6x – 2y = 1 and x2 + y2 – 6kx + 4y – 1 = 0 are collinear, is __________.

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Solution

Centre of x2+y2=1 is 0, 0Centre of x2+y2+6x-2y=1 is -3, 1and centre of x2+y2-6kx+4y-1 is 3k, -2 are collinear i.e. 001-3113k-21=0 i.e. area of triangle with three collinear points is zero.i.e. 6-3k=0i.e. 3k=6i.e. k=2

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