If left(0, frac{3}{4}right) is the radical ce | vedclass

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(C) The radical axis of $S$ and $S^{\prime}$ is given by $S-S^{\prime}=0$: $\Rightarrow (x^2+y^2+\alpha x+6y) - (x^2+y^2+2\alpha x+\alpha y+6) = 0$ $\

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$\frac{\sqrt{65}}{4}$

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(C) The radical axis of $S$ and $S^{\prime}$ is given by $S-S^{\prime}=0$: $\Rightarrow (x^2+y^2+\alpha x+6y) - (x^2+y^2+2\alpha x+\alpha y+6) = 0$ $\Rightarrow -\alpha x + (6-\alpha)y - 6 = 0$ Since $\left(0, \frac{3}{4}\right)$ lies on this axis: $-\alpha(0) + (6-\alpha)(\frac{3}{4}) - 6 = 0$ $\Rightarrow \frac{18-3\alpha}{4} = 6$ $\Rightarrow 18-3\alpha = 24$ $\Rightarrow -3\alpha = 6$ $\Rightarrow \alpha = -2$. Now,the circle $S^{\prime}$ is $x^2+y^2+2(-2)x+(-2)y+6 = 0$,which is $x^2+y^2-4x-2y+6 = 0$. The centre of $S^{\prime}$ is $C = (-g, -f) = (2, 1)$. The radical centre is $P = \left(0, \frac{3}{4}\right)$. The distance $PC = \sqrt{(2-0)^2 + (1-\frac{3}{4})^2} = \sqrt{4 + (\frac{1}{4})^2} = \sqrt{4 + \frac{1}{16}} = \sqrt{\frac{65}{16}} = \frac{\sqrt{65}}{4}$.

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