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

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(C) Let the equations of the circles be:$S_1: x^2+y^2+2x+3y+1=0$$S_2: x^2+y^2+x-y+3=0$$S_3: x^2+y^2-3x+2y+5=0$The radical axis of $S_1$ and $S_2$ is g

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$\left(\frac{14}{19}, \frac{6}{19}\right)$

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(C) Let the equations of the circles be:$S_1: x^2+y^2+2x+3y+1=0$$S_2: x^2+y^2+x-y+3=0$$S_3: x^2+y^2-3x+2y+5=0$The radical axis of $S_1$ and $S_2$ is given by $S_1 - S_2 = 0$:$(x^2+y^2+2x+3y+1) - (x^2+y^2+x-y+3) = 0$$x + 4y - 2 = 0 \quad \dots (i)$The radical axis of $S_2$ and $S_3$ is given by $S_2 - S_3 = 0$:$(x^2+y^2+x-y+3) - (x^2+y^2-3x+2y+5) = 0$$4x - 3y - 2 = 0 \quad \dots (ii)$Solving equations $(i)$ and $(ii)$ for $x$ and $y$:From $(i)$,$x = 2 - 4y$. Substituting into $(ii)$:$4(2 - 4y) - 3y - 2 = 0$$8 - 16y - 3y - 2 = 0$$6 - 19y = 0 \Rightarrow y = \frac{6}{19}$Substituting $y = \frac{6}{19}$ into $x = 2 - 4y$:$x = 2 - 4\left(\frac{6}{19}\right) = 2 - \frac{24}{19} = \frac{38-24}{19} = \frac{14}{19}$Thus,the radical centre is $\left(\frac{14}{19}, \frac{6}{19}\right)$.

The radical centre of the circles $x^2+y^2+2x+3y+1=0$,$x^2+y^2+x-y+3=0$,and $x^2+y^2-3x+2y+5=0$ is

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