The length of the transverse common tangent o | vedclass

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(D) For circle $S_1: x^2+y^2-2x+4y+4=0$,the centre is $C_1(1, -2)$ and radius $r_1 = \sqrt{1^2 + (-2)^2 - 4} = 1$. For circle $S_2: x^2+y^2+4x-2y+1=0$

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$3$

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(D) For circle $S_1: x^2+y^2-2x+4y+4=0$,the centre is $C_1(1, -2)$ and radius $r_1 = \sqrt{1^2 + (-2)^2 - 4} = 1$. For circle $S_2: x^2+y^2+4x-2y+1=0$,the centre is $C_2(-2, 1)$ and radius $r_2 = \sqrt{(-2)^2 + 1^2 - 1} = 2$. The distance between the centres $d$ is given by $d = \sqrt{(1 - (-2))^2 + (-2 - 1)^2} = \sqrt{3^2 + (-3)^2} = \sqrt{18} = 3\sqrt{2}$. The length of the transverse common tangent $L$ is given by the formula $L = \sqrt{d^2 - (r_1 + r_2)^2}$. Substituting the values,$L = \sqrt{(3\sqrt{2})^2 - (1 + 2)^2} = \sqrt{18 - 9} = \sqrt{9} = 3$ units.

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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