The equation of the tangent at the point (0, | vedclass

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(A) Let the equation of the circle be $x^2 + y^2 + 2 g x + 2 f y + c = 0$. Since it cuts the given circles orthogonally,we use the condition $2 g g_1

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$y = 3$

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(A) Let the equation of the circle be $x^2 + y^2 + 2 g x + 2 f y + c = 0$. Since it cuts the given circles orthogonally,we use the condition $2 g g_1 + 2 f f_1 = c + c_1$. For $S_1: x^2 + y^2 - 2 x + 6 y = 0$,we have $-2 g + 6 f = c$. For $S_2: x^2 + y^2 - 4 x - 2 y + 6 = 0$,we have $-4 g - 2 f = c + 6$. For $S_3: x^2 + y^2 - 12 x + 2 y + 3 = 0$,we have $-12 g + 2 f = c + 3$. Solving these equations,we get $g = 0$,$f = -3/4$,and $c = -9/2$. The circle equation is $x^2 + y^2 - 3/2 y - 9/2 = 0$. The equation of the tangent at $(0, 3)$ is given by $x x_1 + y y_1 + g(x + x_1) + f(y + y_1) + c = 0$. Substituting $(x_1, y_1) = (0, 3)$,$g = 0$,$f = -3/4$,and $c = -9/2$: $x(0) + y(3) + 0(x + 0) - 3/4(y + 3) - 9/2 = 0$. $3 y - 3/4 y - 9/4 - 18/4 = 0$. $9/4 y = 27/4$. $y = 3$.

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

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