If the circle $3 x^{2} + 3 y^{2} + 10 x + y - 27 = 0$ bisects the circumference of the circle $x^{2} + y^{2} = k$ then $k^{2} - 1 =$

27

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728

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9

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80

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Solution

The correct option is D $80$
$3 x^{2} + 3 y^{2} + 10 x + y - 27 = 0$ bisects the circumference of
$x^{2} + y^{2} = k .$ means.
common chord of both circle paff. through centre of $x^{2} + y^{2} = a^{2} .$
$∴ \frac{10 x}{3} + \frac{y}{3} - 9 + k = 0.$ is common chord of both circles
$10 x + y + 3 (k - 9) = 0$
$∴ (0, 0)$ lie on $10 x + y + 3 (k - 9) = 0.$
$k = 9$
$\Rightarrow k^{2} - 1 = 81 - 1 = 80$

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