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(C) The given circle equation is $x^2+y^2-4x+6y-12=0$. Comparing this with the general form $x^2+y^2+2gx+2fy+c=0$,we get $g=-2$,$f=3$,and $c=-12$. The

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$5\sqrt{3}$

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(C) The given circle equation is $x^2+y^2-4x+6y-12=0$. Comparing this with the general form $x^2+y^2+2gx+2fy+c=0$,we get $g=-2$,$f=3$,and $c=-12$. The centre of this circle is $C_1 = (-g, -f) = (2, -3)$ and its radius $r_1$ is $\sqrt{g^2+f^2-c} = \sqrt{(-2)^2+3^2-(-12)} = \sqrt{4+9+12} = \sqrt{25} = 5$. Since the diameter of this circle is a chord of circle $S$,the length of this chord is equal to the diameter of the first circle,which is $2r_1 = 2(5) = 10$. Let the centre of circle $S$ be $C_2 = (-3, 2)$. The perpendicular distance $d$ from $C_2$ to the chord is the distance between $(2, -3)$ and $(-3, 2)$. $d = \sqrt{(-3-2)^2 + (2-(-3))^2} = \sqrt{(-5)^2 + 5^2} = \sqrt{25+25} = \sqrt{50} = 5\sqrt{2}$. In circle $S$,the radius $R$,the perpendicular distance $d$,and half the chord length $a$ form a right-angled triangle,where $a = \frac{10}{2} = 5$. Using Pythagoras theorem,$R^2 = d^2 + a^2 = (5\sqrt{2})^2 + 5^2 = 50 + 25 = 75$. Therefore,$R = \sqrt{75} = 5\sqrt{3}$.

If one of the diameters of the circle,given by the equation $x^2+y^2-4x+6y-12=0$,is a chord of a circle,$S$,whose centre is at $(-3,2)$,then the length of the radius of $S$ is . . . . . . units.

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