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

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(D) The equations of the given 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$The radical axis of $S_1$ and $S_2$ i

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$\left(\frac{7}{6}, \frac{11}{6}\right)$

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(D) The equations of the given 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$The radical axis of $S_1$ and $S_2$ is given by $S_1 - S_2 = 0$:$(x^2+y^2-4x-6y+5) - (x^2+y^2-2x-4y-1) = 0$$-2x - 2y + 6 = 0 \Rightarrow x+y=3$ ... $(i)$The radical axis of $S_2$ and $S_3$ is given by $S_2 - S_3 = 0$:$(x^2+y^2-2x-4y-1) - (x^2+y^2-6x-2y) = 0$$4x - 2y - 1 = 0$ ... (ii)To find the radical centre,solve equations $(i)$ and (ii):From $(i)$,$y = 3 - x$. Substitute into (ii):$4x - 2(3 - x) - 1 = 0$$4x - 6 + 2x - 1 = 0$$6x = 7 \Rightarrow x = \frac{7}{6}$Substituting $x = \frac{7}{6}$ into $(i)$:$y = 3 - \frac{7}{6} = \frac{18-7}{6} = \frac{11}{6}$Thus,the radical centre is $\left(\frac{7}{6}, \frac{11}{6}\right)$.

The radical centre of the circles $x^2+y^2-4x-6y+5=0$,$x^2+y^2-2x-4y-1=0$ and $x^2+y^2-6x-2y=0$ is equal to

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