The ratio of the areas of the concentric circ | vedclass

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(C) The general equation of a circle is $x^2+y^2+2gx+2fy+c=0$,where the radius $r = \sqrt{g^2+f^2-c}$.For the first circle $C_1: x^2+y^2-6x+12y+15=0$,

Vedclass · Accepted Answer

$1: 2$

Vedclass · Answer

(C) The general equation of a circle is $x^2+y^2+2gx+2fy+c=0$,where the radius $r = \sqrt{g^2+f^2-c}$.For the first circle $C_1: x^2+y^2-6x+12y+15=0$,we have $g=-3, f=6, c=15$.$r_1 = \sqrt{(-3)^2+6^2-15} = \sqrt{9+36-15} = \sqrt{30}$.For the second circle $C_2: x^2+y^2-6x+12y-15=0$,we have $g=-3, f=6, c=-15$.$r_2 = \sqrt{(-3)^2+6^2-(-15)} = \sqrt{9+36+15} = \sqrt{60}$.The ratio of the areas is $\frac{\pi r_1^2}{\pi r_2^2} = \frac{r_1^2}{r_2^2} = \frac{30}{60} = \frac{1}{2}$.Thus,the ratio is $1:2$.

The ratio of the areas of the concentric circles $x^2+y^2-6x+12y+15=0$ and $x^2+y^2-6x+12y-15=0$ is

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