The radical centre of the circles x^2+y^2-4x- | vedclass

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(D) The equations of the circles are:$S_1: x^2+y^2-4x-6y+5=0$$S_2: x^2+y^2-2x-4y-1=0$$S_3: x^2+y^2-6x-2y=0$To find the radical centre,we find the radi

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$18x-12y+1=0$

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(D) The equations of the circles are:$S_1: x^2+y^2-4x-6y+5=0$$S_2: x^2+y^2-2x-4y-1=0$$S_3: x^2+y^2-6x-2y=0$To find the radical centre,we find the radical axes by subtracting the equations:$S_1 - S_2 = 0$ $\Rightarrow (x^2+y^2-4x-6y+5) - (x^2+y^2-2x-4y-1) = 0$ $\Rightarrow -2x-2y+6=0$ $\Rightarrow x+y-3=0$ (Equation $i$)$S_2 - S_3 = 0$ $\Rightarrow (x^2+y^2-2x-4y-1) - (x^2+y^2-6x-2y) = 0$ $\Rightarrow 4x-2y-1=0$ (Equation $ii$)Solving equations $(i)$ and $(ii)$:From $(i)$,$y = 3-x$.Substitute into $(ii)$: $4x - 2(3-x) - 1 = 0$ $\Rightarrow 4x - 6 + 2x - 1 = 0$ $\Rightarrow 6x = 7$ $\Rightarrow x = \frac{7}{6}$.Then $y = 3 - \frac{7}{6} = \frac{11}{6}$.The radical centre is $(\frac{7}{6}, \frac{11}{6})$.Checking the options,for option $D$: $18(\frac{7}{6}) - 12(\frac{11}{6}) + 1 = 3(7) - 2(11) + 1 = 21 - 22 + 1 = 0$.Thus,the radical centre lies on the line $18x-12y+1=0$.

The radical centre of the circles $x^2+y^2-4x-6y+5=0$,$x^2+y^2-2x-4y-1=0$ and $x^2+y^2-6x-2y=0$ lies on the line

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