If theta is the angle between the circles x^2 | vedclass

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(A) The given circles are $S_1: x^2+y^2-2x-4y-4=0$ and $S_2: x^2+y^2-8x-12y+43=0$. Comparing with $x^2+y^2+2gx+2fy+c=0$,we have: For $S_1: g_1 = -1, f

Vedclass · Accepted Answer

$11$

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(A) The given circles are $S_1: x^2+y^2-2x-4y-4=0$ and $S_2: x^2+y^2-8x-12y+43=0$. Comparing with $x^2+y^2+2gx+2fy+c=0$,we have: For $S_1: g_1 = -1, f_1 = -2, c_1 = -4$. Radius $r_1 = \sqrt{1+4-(-4)} = 3$. For $S_2: g_2 = -4, f_2 = -6, c_2 = 43$. Radius $r_2 = \sqrt{16+36-43} = 3$. The distance between centers $C_1(1, 2)$ and $C_2(4, 6)$ is $d = \sqrt{(4-1)^2 + (6-2)^2} = \sqrt{3^2+4^2} = 5$. The angle $	heta$ between two circles is given by $\cos 	heta = \frac{d^2 - r_1^2 - r_2^2}{2r_1r_2}$. Substituting the values: $\cos 	heta = \frac{5^2 - 3^2 - 3^2}{2(3)(3)} = \frac{25 - 9 - 9}{18} = \frac{7}{18}$. Thus,$\sec 	heta = \frac{18}{7}$. Now,$|7 \sec 	heta - 18 \cos 	heta| = |7(\frac{18}{7}) - 18(\frac{7}{18})| = |18 - 7| = 11$.

If $\theta$ is the angle between the circles $x^2+y^2-2x-4y-4=0$ and $x^2+y^2-8x-12y+43=0$,then $|7 \sec \theta - 18 \cos \theta| = $

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