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(D) Given circle: $x^2 + y^2 + 14x + 6y + 2 = 0$. Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = 7$,$f = 3$,and $c = 2$. Let the required

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$x^2 + y^2 - 4y - 14 = 0$

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(D) Given circle: $x^2 + y^2 + 14x + 6y + 2 = 0$. Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $g = 7$,$f = 3$,and $c = 2$. Let the required circle be $x^2 + y^2 + 2g'x + 2f'y + c' = 0$. The centre of this circle is $(-g', -f') = (0, 2)$,so $g' = 0$ and $f' = -2$. Since the circles intersect orthogonally,the condition is $2gg' + 2ff' = c + c'$. Substituting the values: $2(7)(0) + 2(3)(-2) = 2 + c'$. $0 - 12 = 2 + c' \Rightarrow c' = -14$. Substituting $g'$,$f'$,and $c'$ into the general equation: $x^2 + y^2 + 2(0)x + 2(-2)y - 14 = 0$. Thus,the equation is $x^2 + y^2 - 4y - 14 = 0$.

The equation of a circle that intersects the circle $x^2 + y^2 + 14x + 6y + 2 = 0$ orthogonally and whose centre is $(0, 2)$ is

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