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(C) The radical axis of two circles $S_1=0$ and $S_2=0$ is given by $S_1-S_2=0$. Given $S_1: x^2+y^2-5x+6y+12=0$ and $S_2: x^2+y^2+6x-4y-14=0$. Subtra

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$10x+11y-21=0$

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(C) The radical axis of two circles $S_1=0$ and $S_2=0$ is given by $S_1-S_2=0$. Given $S_1: x^2+y^2-5x+6y+12=0$ and $S_2: x^2+y^2+6x-4y-14=0$. Subtracting $S_2$ from $S_1$: $(x^2+y^2-5x+6y+12) - (x^2+y^2+6x-4y-14) = 0$ $-11x+10y+26=0$ or $11x-10y-26=0$. The slope of the radical axis is $m_1 = \frac{11}{10}$. The slope of the line perpendicular to the radical axis is $m_2 = -\frac{1}{m_1} = -\frac{10}{11}$. The equation of the line passing through $(1,1)$ with slope $m_2 = -\frac{10}{11}$ is: $y-1 = -\frac{10}{11}(x-1)$ $11(y-1) = -10(x-1)$ $11y-11 = -10x+10$ $10x+11y-21=0$.

The equation of the line perpendicular to the radical axis of two circles $x^2+y^2-5x+6y+12=0$ and $x^2+y^2+6x-4y-14=0$,and passing through $(1,1)$ is:

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