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(D) The given equations of the circles are:$x^2 + y^2 - 6x + 4y + 9 = 0 \Rightarrow (x-3)^2 + (y+2)^2 = 2^2 \quad \dots(i)$$x^2 + y^2 + 2x - 2y + 1 =

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$2\sqrt{6}$

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(D) The given equations of the circles are:$x^2 + y^2 - 6x + 4y + 9 = 0 \Rightarrow (x-3)^2 + (y+2)^2 = 2^2 \quad \dots(i)$$x^2 + y^2 + 2x - 2y + 1 = 0 \Rightarrow (x+1)^2 + (y-1)^2 = 1^2 \quad \dots(ii)$From $(i)$ and $(ii)$,the centers and radii are:$C_1 = (3, -2), r_1 = 2$$C_2 = (-1, 1), r_2 = 1$Calculate the distance between the centers $C_1$ and $C_2$:$C_1C_2 = \sqrt{(3 - (-1))^2 + (-2 - 1)^2} = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = 5$The length of the direct common tangent $AB$ is given by the formula:$AB = \sqrt{(C_1C_2)^2 - (r_1 - r_2)^2}$$AB = \sqrt{5^2 - (2 - 1)^2} = \sqrt{25 - 1} = \sqrt{24} = 2\sqrt{6}$

If a direct common tangent drawn to the circles $x^2+y^2-6x+4y+9=0$ and $x^2+y^2+2x-2y+1=0$ touches the circles at $A$ and $B$,then $AB=$

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