If the circle S=0 cuts the circles x^2+y^2-2x | vedclass

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(B) Let the circle $S$ be $x^2+y^2+2gx+2fy+c=0$.
Since $S$ cuts the given circles orthogonally,we use the condition $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.

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$y=3$

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(B) Let the circle $S$ be $x^2+y^2+2gx+2fy+c=0$.
Since $S$ cuts the given circles orthogonally,we use the condition $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.
For $S_1: x^2+y^2-2x+6y=0$,we have $g_1=-1, f_1=3, c_1=0$. Thus,$2g(-1) + 2f(3) = c + 0 \Rightarrow -2g + 6f = c$ $(i)$.
For $S_2: x^2+y^2-4x-2y+6=0$,we have $g_2=-2, f_2=-1, c_2=6$. Thus,$2g(-2) + 2f(-1) = c + 6 \Rightarrow -4g - 2f = c + 6$ $(ii)$.
For $S_3: x^2+y^2-12x+2y+3=0$,we have $g_3=-6, f_3=1, c_3=3$. Thus,$2g(-6) + 2f(1) = c + 3 \Rightarrow -12g + 2f = c + 3$ $(iii)$.
Subtracting $(ii)$ from $(i)$: $2g + 8f = -6 \Rightarrow g + 4f = -3$ $(iv)$.
Subtracting $(iii)$ from $(ii)$: $8g - 4f = 3$ $(v)$.
Adding $(iv)$ and $(v)$: $9g = 0 \Rightarrow g = 0$. Then $4f = -3 \Rightarrow f = -3/4$.
Substituting into $(i)$: $c = -2(0) + 6(-3/4) = -18/4 = -9/2$.
The circle $S$ is $x^2+y^2 - \frac{3}{2}y - \frac{9}{2} = 0$.
The tangent at $(0,3)$ is given by $xx_1 + yy_1 + g(x+x_1) + f(y+y_1) + c = 0$.
Substituting $(0,3)$: $x(0) + y(3) + 0(x+0) - \frac{3}{4}(y+3) - \frac{9}{2} = 0$.
$3y - \frac{3}{4}y - \frac{9}{4} - \frac{18}{4} = 0 \Rightarrow \frac{9}{4}y - \frac{27}{4} = 0 \Rightarrow 9y = 27 \Rightarrow y = 3$.

If the circle $S=0$ cuts the circles $x^2+y^2-2x+6y=0$,$x^2+y^2-4x-2y+6=0$,and $x^2+y^2-12x+2y+3=0$ orthogonally,then the equation of the tangent at $(0,3)$ on $S=0$ is

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