If the circles x^2+y^2+2 alpha x+2 y-8=0 and | vedclass

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(C) The given equations of the circles are $S_1: x^2+y^2+2 \alpha x+2 y-8=0$ and $S_2: x^2+y^2-2 x+\alpha y-14=0$. Comparing with $x^2+y^2+2gx+2fy+c=0

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$\sqrt{629}$

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(C) The given equations of the circles are $S_1: x^2+y^2+2 \alpha x+2 y-8=0$ and $S_2: x^2+y^2-2 x+\alpha y-14=0$. Comparing with $x^2+y^2+2gx+2fy+c=0$,we get $(g_1, f_1, c_1) = (\alpha, 1, -8)$ and $(g_2, f_2, c_2) = (-1, \frac{\alpha}{2}, -14)$. Since the circles intersect orthogonally,the condition is $2g_1g_2 + 2f_1f_2 = c_1 + c_2$. Substituting the values: $2(\alpha)(-1) + 2(1)(\frac{\alpha}{2}) = -8 - 14$. $-2\alpha + \alpha = -22$,which gives $\alpha = 22$. The centre of $S_1$ is $C_1 = (-g_1, -f_1) = (-22, -1)$. The centre of $S_2$ is $C_2 = (-g_2, -f_2) = (1, -\frac{22}{2}) = (1, -11)$. The distance between the centres $C_1$ and $C_2$ is $d = \sqrt{(1 - (-22))^2 + (-11 - (-1))^2} = \sqrt{(23)^2 + (-10)^2} = \sqrt{529 + 100} = \sqrt{629}$.

If the circles $x^2+y^2+2 \alpha x+2 y-8=0$ and $x^2+y^2-2 x+\alpha y-14=0$ intersect orthogonally,then the distance between their centres is

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