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

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(A) Let the equations of the circles be:$S_1: x^2+y^2-1=0$ ...$(i)$$S_2: x^2+y^2-8x+15=0$ ...(ii)$S_3: x^2+y^2+10y+24=0$ ...(iii)The radical axis of $

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$\left(2, -\frac{5}{2}\right)$

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(A) Let the equations of the circles be:$S_1: x^2+y^2-1=0$ ...$(i)$$S_2: x^2+y^2-8x+15=0$ ...(ii)$S_3: x^2+y^2+10y+24=0$ ...(iii)The radical axis of $S_1$ and $S_2$ is given by $S_1 - S_2 = 0$:$(x^2+y^2-1) - (x^2+y^2-8x+15) = 0$$8x - 16 = 0 \Rightarrow x = 2$The radical axis of $S_1$ and $S_3$ is given by $S_1 - S_3 = 0$:$(x^2+y^2-1) - (x^2+y^2+10y+24) = 0$$-10y - 25 = 0$ $\Rightarrow 10y = -25$ $\Rightarrow y = -\frac{5}{2}$The radical centre is the intersection of these radical axes,which is $\left(2, -\frac{5}{2}\right)$.

The radical centre of the three circles $x^2+y^2-1=0$,$x^2+y^2-8x+15=0$ and $x^2+y^2+10y+24=0$ is

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