If the circle S = x^2 + y^2 + 2gx + 2fy + c = | vedclass

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(B) The condition for two circles $x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$ and $x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$ to be orthogonal is $2g_1g_2 + 2f_1f

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$\sqrt{3}x + 2y - 7 = 0$

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(B) The condition for two circles $x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$ and $x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$ to be orthogonal is $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.Applying this to the given circles:$1$) $2g(2) + 2f(2) = c + 7 \Rightarrow 4g + 4f = c + 7$$2$) $2g(-2) + 2f(2) = c + 7 \Rightarrow -4g + 4f = c + 7$$3$) $2g(-2) + 2f(-2) = c + 7 \Rightarrow -4g - 4f = c + 7$Subtracting $(2)$ from $(1)$: $8g = 0 \Rightarrow g = 0$.Adding $(2)$ and $(3)$: $-8f = 2(c + 7) \Rightarrow -4f = c + 7$. Since $4g + 4f = c + 7$ and $g=0$,we have $4f = c + 7$. Thus,$c + 7 = -(c + 7) \Rightarrow c = -7$.Substituting $c = -7$ into $4f = c + 7$,we get $4f = 0 \Rightarrow f = 0$.The circle equation is $S: x^2 + y^2 - 7 = 0$.The equation of the tangent at $(x_1, y_1)$ is $xx_1 + yy_1 + c = 0$.At $(\sqrt{3}, 2)$,the tangent is $\sqrt{3}x + 2y - 7 = 0$.

If the circle $S = x^2 + y^2 + 2gx + 2fy + c = 0$ cuts each of the three circles $x^2 + y^2 + 4x + 4y + 7 = 0$,$x^2 + y^2 - 4x + 4y + 7 = 0$,and $x^2 + y^2 - 4x - 4y + 7 = 0$ orthogonally,then the equation of the tangent drawn at the point $(\sqrt{3}, 2)$ to the circle $S = 0$ is

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