x^2+y^2+2x-6y-6=0 and x^2+y^2-6x-2y+k=0 are t | vedclass

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(B) For the circle $x^2+y^2+2x-6y-6=0$,the centre $C_1 = (-1, 3)$ and radius $r_1 = \sqrt{(-1)^2 + 3^2 - (-6)} = \sqrt{1+9+6} = 4$. For the circle $x^

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$\frac{74}{9}$

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(B) For the circle $x^2+y^2+2x-6y-6=0$,the centre $C_1 = (-1, 3)$ and radius $r_1 = \sqrt{(-1)^2 + 3^2 - (-6)} = \sqrt{1+9+6} = 4$. For the circle $x^2+y^2-6x-2y+k=0$,the centre $C_2 = (3, 1)$ and radius $r_2 = \sqrt{3^2 + 1^2 - k} = \sqrt{10-k}$. The distance between the centres $d^2 = C_1C_2^2 = (3 - (-1))^2 + (1 - 3)^2 = 4^2 + (-2)^2 = 16 + 4 = 20$. Using the law of cosines for the triangle formed by the two centres and an intersection point,$d^2 = r_1^2 + r_2^2 - 2r_1r_2 \cos 	heta$. Substituting the values: $20 = 16 + (10-k) - 2(4)(\sqrt{10-k})(\frac{-5}{24})$. $20 = 26 - k + \frac{5}{3}\sqrt{10-k}$. $k - 6 = \frac{5}{3}\sqrt{10-k}$. Squaring both sides: $(k-6)^2 = \frac{25}{9}(10-k)$. $9(k^2 - 12k + 36) = 250 - 25k$. $9k^2 - 108k + 324 = 250 - 25k$. $9k^2 - 83k + 74 = 0$. $(k-1)(9k-74) = 0$. Since $k$ is not an integer,$k = \frac{74}{9}$.

$x^2+y^2+2x-6y-6=0$ and $x^2+y^2-6x-2y+k=0$ are two intersecting circles and $k$ is not an integer. If $\theta$ is the angle between the two circles and $\cos \theta = \frac{-5}{24}$,then $k=$

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