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If the poles of the line $x-y=0$ with respect to the circles $x^2+y^2-2 g_i x+c_i^2=0(i=1,2,3)$ are $\left(\alpha_i, \beta_i\right)$, then $\sum_{i=1}^3 \frac{\alpha_i+\beta_i}{g_i}=$
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Since equation polar of point $\left(\alpha_i, \beta_i\right)$ with respect to the circle $x^2+y^2-2 g_i x+c_i^2=0$ is
$\alpha_i x+\beta_i y-g_i\left(x+\alpha_i\right)+c_i^2=0$
$\Rightarrow \quad\left(\alpha_i-g_i\right) x+\beta_i y+c_i^2-\alpha_i g_i=0$
Now, on comparing with line $x-y=0$, we get
$\frac{\alpha_i-g_i}{1}=\frac{\beta_i}{-1}=\frac{c_i^2-\alpha_i g_i}{0}$
$\Rightarrow \quad c_i^2=\alpha_i g_i$ and $\alpha_i+\beta_i=g_i$
$\therefore \quad \sum_{i=1}^3 \frac{\alpha_i+\beta_i}{g_i}=\sum_{i=1}^3(\mathrm{l})=3$
$\alpha_i x+\beta_i y-g_i\left(x+\alpha_i\right)+c_i^2=0$
$\Rightarrow \quad\left(\alpha_i-g_i\right) x+\beta_i y+c_i^2-\alpha_i g_i=0$
Now, on comparing with line $x-y=0$, we get
$\frac{\alpha_i-g_i}{1}=\frac{\beta_i}{-1}=\frac{c_i^2-\alpha_i g_i}{0}$
$\Rightarrow \quad c_i^2=\alpha_i g_i$ and $\alpha_i+\beta_i=g_i$
$\therefore \quad \sum_{i=1}^3 \frac{\alpha_i+\beta_i}{g_i}=\sum_{i=1}^3(\mathrm{l})=3$
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