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A circle has centre on the line y=1 and cuts orthogonally the circle $x^{2} + y^{2} + 6 y + 3 = 0$ then it always passes through two fixed points, one of them is

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a

$(0, 1 + \sqrt{10})$

b

(1,0)

c

(0,1)

d

(0,0)

answer is A.

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Detailed Solution

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$\begin{array}{l} P (k, 1) \\ {(x - k)}^{2} + {(y - (1))}^{2} = k^{2} + {(1)}^{2} + 6 (1) + 3 \\ \Rightarrow x^{2} + y^{2} - 2 k x - 2 (1) y = 9 \Rightarrow x^{2} + y^{2} - 2 y - 9 - 2 k x = 0 \\ y = \frac{2 \pm \sqrt{40}}{2} s o l v i n g (0, 1 + \sqrt{10}), (0, 1 - \sqrt{10}) y = 1 \pm \sqrt{10} \end{array}$

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