If the angle between the circles x^2+y^2-4x-6 | vedclass

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(A) Given circles are $S_1: x^2+y^2-4x-6y+k=0$ and $S_2: x^2+y^2+8x-4y+11=0$.For $S_1$,centre $C_1(2, 3)$ and radius $r_1 = \sqrt{2^2+3^2-k} = \sqrt{1

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$-36$

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(A) Given circles are $S_1: x^2+y^2-4x-6y+k=0$ and $S_2: x^2+y^2+8x-4y+11=0$.For $S_1$,centre $C_1(2, 3)$ and radius $r_1 = \sqrt{2^2+3^2-k} = \sqrt{13-k}$.For $S_2$,centre $C_2(-4, 2)$ and radius $r_2 = \sqrt{(-4)^2+2^2-11} = \sqrt{16+4-11} = \sqrt{9} = 3$.The distance between centres $d = C_1C_2 = \sqrt{(2 - (-4))^2 + (3 - 2)^2} = \sqrt{6^2 + 1^2} = \sqrt{37}$.The angle $	heta$ between two circles is given by $\cos 	heta = \left| \frac{r_1^2 + r_2^2 - d^2}{2r_1r_2} ight|$.Given $	heta = \frac{\pi}{3}$,so $\cos \frac{\pi}{3} = \frac{1}{2}$.$\frac{1}{2} = \left| \frac{(13-k) + 9 - 37}{2 \cdot \sqrt{13-k} \cdot 3} ight| = \left| \frac{-15-k}{6\sqrt{13-k}} ight|$.$3\sqrt{13-k} = |-(15+k)| = |15+k|$.Squaring both sides: $9(13-k) = (15+k)^2$.$117 - 9k = 225 + 30k + k^2$.$k^2 + 39k + 108 = 0$.$(k+36)(k+3) = 0$.So $k = -36$ or $k = -3$.Checking the condition $3\sqrt{13-k} = |15+k|$:If $k = -3$,$3\sqrt{16} = 12$ and $|15-3| = 12$. (Valid)If $k = -36$,$3\sqrt{13+36} = 3(7) = 21$ and $|15-36| = |-21| = 21$. (Valid)Since $-36$ is an option,the answer is $-36$.

If the angle between the circles $x^2+y^2-4x-6y+k=0$ and $x^2+y^2+8x-4y+11=0$ is $\frac{\pi}{3}$,then a value of $k$ is

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