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

$Centre of x^{2} + y^{2} = 1 is (0, 0) Centre of x^{2} + y^{2} + 6 x - 2 y = 1 is (- 3, 1) and centre of x^{2} + y^{2} - 6 k x + 4 y - 1 is (3 k, - 2) are collinear i . e . |\begin{matrix} 0 & 0 & 1 \\ - 3 & 1 & 1 \\ 3 k & - 2 & 1 \end{matrix}| = 0 (i . e . area of triangle with three collinear points is zero .) i . e . (6 - 3 k) = 0 i . e . 3 k = 6 i . e . k = 2$

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x^{2} + y^{2} = 1,

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