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(C) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.Since it intersects the circles $x^2 + y^2 + 2x + 4y + 6 = 0$ and $x^2 + y^2 + 4

Vedclass · Accepted Answer

$x^2 + y^2 - 16x + 12y + 2 = 0$

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(C) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.Since it intersects the circles $x^2 + y^2 + 2x + 4y + 6 = 0$ and $x^2 + y^2 + 4x + 6y + 2 = 0$ orthogonally,we use the condition $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.For the first circle: $2g(1) + 2f(2) = c + 6 \implies 2g + 4f - c = 6$.For the second circle: $2g(2) + 2f(3) = c + 2 \implies 4g + 6f - c = 2$.Since the circle passes through $(1, 1)$,we have $1^2 + 1^2 + 2g(1) + 2f(1) + c = 0 \implies 2g + 2f + c = -2$.Solving the system of equations:$2g + 4f - c = 6$$4g + 6f - c = 2$$2g + 2f + c = -2$Subtracting the first from the second: $2g + 2f = -4 \implies g + f = -2$.Adding the first and third: $4g + 6f = 4 \implies 2g + 3f = 2$.Solving these gives $f = 6$ and $g = -8$.Substituting into the third equation: $2(-8) + 2(6) + c = -2 \implies -16 + 12 + c = -2 \implies c = 2$.Thus,the equation is $x^2 + y^2 - 16x + 12y + 2 = 0$.

The equation of the circle which passes through the point $(1, 1)$ and intersects the given circles $x^2 + y^2 + 2x + 4y + 6 = 0$ and $x^2 + y^2 + 4x + 6y + 2 = 0$ orthogonally is:

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