The length of the transverse common tangent o | vedclass

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(D) For the circle $x^2+y^2-2x+4y+4=0$,the center $C_1 = (1, -2)$ and radius $r_1 = \sqrt{1^2 + (-2)^2 - 4} = \sqrt{1+4-4} = 1$. For the circle $x^2+y

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$3$

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(D) For the circle $x^2+y^2-2x+4y+4=0$,the center $C_1 = (1, -2)$ and radius $r_1 = \sqrt{1^2 + (-2)^2 - 4} = \sqrt{1+4-4} = 1$. For the circle $x^2+y^2+4x-2y+1=0$,the center $C_2 = (-2, 1)$ and radius $r_2 = \sqrt{(-2)^2 + 1^2 - 1} = \sqrt{4+1-1} = 2$. The distance between the centers $C_1$ and $C_2$ is $d = \sqrt{(-2-1)^2 + (1-(-2))^2} = \sqrt{(-3)^2 + 3^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt{2}$. The length of the transverse common tangent is given by the formula $L = \sqrt{d^2 - (r_1+r_2)^2}$. Substituting the values,$L = \sqrt{(\sqrt{18})^2 - (1+2)^2} = \sqrt{18 - 3^2} = \sqrt{18 - 9} = \sqrt{9} = 3$.

The length of the transverse common tangent of the circles $x^2+y^2-2x+4y+4=0$ and $x^2+y^2+4x-2y+1=0$ is

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