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(A) The equation of the polar of a point $(\alpha_i, \beta_i)$ with respect to the circle $x^2+y^2-2g_ix+c_i^2=0$ is given by $\alpha_ix + \beta_iy -

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$3$

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(A) The equation of the polar of a point $(\alpha_i, \beta_i)$ with respect to the circle $x^2+y^2-2g_ix+c_i^2=0$ is given by $\alpha_ix + \beta_iy - g_i(x+\alpha_i) + c_i^2 = 0$. Rearranging the terms,we get $(\alpha_i - g_i)x + \beta_iy + (c_i^2 - \alpha_ig_i) = 0$. Comparing this with the given line $x - y = 0$,we have the ratios of coefficients equal: $\frac{\alpha_i - g_i}{1} = \frac{\beta_i}{-1} = \frac{c_i^2 - \alpha_ig_i}{0}$. From the third part,we get $c_i^2 - \alpha_ig_i = 0$,which implies $c_i^2 = \alpha_ig_i$. From the first two parts,$\alpha_i - g_i = -\beta_i$,which implies $\alpha_i + \beta_i = g_i$. Therefore,$\frac{\alpha_i + \beta_i}{g_i} = \frac{g_i}{g_i} = 1$. Summing this for $i=1, 2, 3$,we get $\sum_{i=1}^3 \frac{\alpha_i + \beta_i}{g_i} = 1 + 1 + 1 = 3$.

If the poles of the line $x-y=0$ with respect to the circles $x^2+y^2-2g_ix+c_i^2=0$ $(i=1, 2, 3)$ are $(\alpha_i, \beta_i)$,then $\sum_{i=1}^3 \frac{\alpha_i+\beta_i}{g_i}=$

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