If P_1, P_2, P_3 are the perimeters of the th | vedclass

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(B) The given circles are:$C_1: x^2+y^2+8x-6y=0$$C_2: 4x^2+4y^2-4x-12y-186=0 \implies x^2+y^2-x-3y-46.5=0$$C_3: x^2+y^2-6x+6y-9=0$Calculating the radi

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$P_1 < P_3 < P_2$

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(B) The given circles are:$C_1: x^2+y^2+8x-6y=0$$C_2: 4x^2+4y^2-4x-12y-186=0 \implies x^2+y^2-x-3y-46.5=0$$C_3: x^2+y^2-6x+6y-9=0$Calculating the radii $r_1, r_2, r_3$:$r_1 = \sqrt{(-4)^2 - (-3)^2 - 0} = \sqrt{16+9} = 5$$r_2 = \sqrt{(0.5)^2 + (1.5)^2 - (-46.5)} = \sqrt{0.25 + 2.25 + 46.5} = \sqrt{49} = 7$$r_3 = \sqrt{(3)^2 + (-3)^2 - (-9)} = \sqrt{9+9+9} = \sqrt{27} = 3\sqrt{3} \approx 5.196$Calculating perimeters $P = 2\pi r$:$P_1 = 2\pi(5) = 10\pi$$P_2 = 2\pi(7) = 14\pi$$P_3 = 2\pi(3\sqrt{3}) = 6\sqrt{3}\pi \approx 10.39\pi$Comparing the values:$10\pi < 10.39\pi < 14\pi \implies P_1 < P_3 < P_2$.

If $P_1, P_2, P_3$ are the perimeters of the three circles $x^2+y^2+8x-6y=0$,$4x^2+4y^2-4x-12y-186=0$,and $x^2+y^2-6x+6y-9=0$ respectively,then

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