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(D) The given circle equation is $x^2 + y^2 + 4x + 16y - 30 = 0$. Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $2g = 4$

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$\sqrt{11}$

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(D) The given circle equation is $x^2 + y^2 + 4x + 16y - 30 = 0$. Comparing this with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$,we get $2g = 4$,$2f = 16$,and $c = -30$. Thus,$g = 2$,$f = 8$,and $c = -30$. The centre of this circle is $C_1 = (-g, -f) = (-2, -8)$. The radius of this circle is $r_1 = \sqrt{g^2 + f^2 - c} = \sqrt{2^2 + 8^2 - (-30)} = \sqrt{4 + 64 + 30} = \sqrt{98}$. Let the other circle have centre $C_2 = (1, 2)$ and radius $r_2$. The distance between the centres $d$ is given by $d^2 = (1 - (-2))^2 + (2 - (-8))^2 = 3^2 + 10^2 = 9 + 100 = 109$. Since the two circles are orthogonal,they satisfy the condition $d^2 = r_1^2 + r_2^2$. Substituting the values,$109 = 98 + r_2^2$. $r_2^2 = 109 - 98 = 11$. Therefore,$r_2 = \sqrt{11}$.

The radius of the circle whose centre lies at $(1, 2)$,while cutting the circle $x^2 + y^2 + 4x + 16y - 30 = 0$ orthogonally,is (in units):

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