The centres of the circles x^2 + y^2 = 1,x^2 | vedclass

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(B) The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$,where the centre is $(-g, -f)$.For the first circle $x^2 + y^2 - 1 = 0$,the ce

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Collinear

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(B) The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$,where the centre is $(-g, -f)$.For the first circle $x^2 + y^2 - 1 = 0$,the centre $C_1$ is $(0, 0)$.For the second circle $x^2 + y^2 + 6x - 2y - 1 = 0$,$2g = 6 \implies g = 3$ and $2f = -2 \implies f = -1$. The centre $C_2$ is $(-3, 1)$.For the third circle $x^2 + y^2 - 12x + 4y - 1 = 0$,$2g = -12 \implies g = -6$ and $2f = 4 \implies f = 2$. The centre $C_3$ is $(6, -2)$.To check if $C_1(0, 0)$,$C_2(-3, 1)$,and $C_3(6, -2)$ are collinear,we check the slope between pairs.Slope of $C_1C_2 = \frac{1 - 0}{-3 - 0} = -\frac{1}{3}$.Slope of $C_2C_3 = \frac{-2 - 1}{6 - (-3)} = \frac{-3}{9} = -\frac{1}{3}$.Since the slopes are equal,the points are collinear.

The centres of the circles $x^2 + y^2 = 1$,$x^2 + y^2 + 6x - 2y = 1$ and $x^2 + y^2 - 12x + 4y = 1$ are

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