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The centre of the circle which circumscribes the square formed by the lines $x^{2} - 8 x + 12 = 0$ and $y^{2} - 14 y + 45 = 0,$ is

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$(3, 7)$

$(4, 7)$

$(2, 5)$

$(6, 9)$

answer is B.

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Detailed Solution

Given $x^{2} - 8 x + 12 = 0$
$\Rightarrow x = 2, x = 6$
Also, $y^{2} - 14 y + 45 = 0$
$\Rightarrow y = 5, y = 9$
The centre of the circle is the centre of the square.
Centre $= (\frac{2 + 6}{2}, \frac{5 + 9}{2}) = (4, 7) .$

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