The angle between the circles x^2+y^2-4x-6y-3 | vedclass

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(A) The given circles are $S_1: x^2+y^2-4x-6y-3=0$ and $S_2: x^2+y^2+8x-4y+11=0$. Comparing with $x^2+y^2+2gx+2fy+c=0$,we get: For $S_1$: $g_1 = -2, f

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$\frac{\pi}{3}$

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(A) The given circles are $S_1: x^2+y^2-4x-6y-3=0$ and $S_2: x^2+y^2+8x-4y+11=0$. Comparing with $x^2+y^2+2gx+2fy+c=0$,we get: For $S_1$: $g_1 = -2, f_1 = -3, c_1 = -3$. For $S_2$: $g_2 = 4, f_2 = -2, c_2 = 11$. The angle $	heta$ between two circles is given by $\cos 	heta = \frac{|2g_1g_2 + 2f_1f_2 - c_1 - c_2|}{2\sqrt{g_1^2+f_1^2-c_1}\sqrt{g_2^2+f_2^2-c_2}}$. Substituting the values: $2g_1g_2 = 2(-2)(4) = -16$. $2f_1f_2 = 2(-3)(-2) = 12$. Numerator: $|-16 + 12 - (-3) - 11| = |-16 + 12 + 3 - 11| = |-12| = 12$. Denominator: $2\sqrt{(-2)^2+(-3)^2-(-3)} \sqrt{4^2+(-2)^2-11} = 2\sqrt{4+9+3}\sqrt{16+4-11} = 2\sqrt{16}\sqrt{9} = 2(4)(3) = 24$. Thus,$\cos 	heta = \frac{12}{24} = \frac{1}{2}$. Therefore,$	heta = \frac{\pi}{3}$.

The angle between the circles $x^2+y^2-4x-6y-3=0$ and $x^2+y^2+8x-4y+11=0$ is

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