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

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(B) The general equation of a circle is $x^2+y^2+2gx+2fy+c=0$,with radius $r = \sqrt{g^2+f^2-c}$.For the first circle $x^2+y^2+8x-6y=0$,$g=4, f=-3, c=

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$P_1 < P_3 < P_2$

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(B) The general equation of a circle is $x^2+y^2+2gx+2fy+c=0$,with radius $r = \sqrt{g^2+f^2-c}$.For the first circle $x^2+y^2+8x-6y=0$,$g=4, f=-3, c=0$. Thus,$r_1 = \sqrt{4^2+(-3)^2-0} = \sqrt{16+9} = 5$. Perimeter $P_1 = 2\pi r_1 = 10\pi$.For the second circle $4x^2+4y^2-4x-12y-186=0$,divide by $4$: $x^2+y^2-x-3y-46.5=0$. Here $g=-0.5, f=-1.5, c=-46.5$. Thus,$r_2 = \sqrt{(-0.5)^2+(-1.5)^2-(-46.5)} = \sqrt{0.25+2.25+46.5} = \sqrt{49} = 7$. Perimeter $P_2 = 2\pi r_2 = 14\pi$.For the third circle $x^2+y^2-6x+6y-9=0$,$g=-3, f=3, c=-9$. Thus,$r_3 = \sqrt{(-3)^2+3^2-(-9)} = \sqrt{9+9+9} = \sqrt{27} = 3\sqrt{3} \approx 3 	imes 1.732 = 5.196$. Perimeter $P_3 = 2\pi r_3 = 6\sqrt{3}\pi \approx 10.39\pi$.Comparing the perimeters: $10\pi < 10.39\pi < 14\pi$,which 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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