The number of common tangents to the circles | vedclass

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

Vedclass · Accepted Answer

$3$

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(B) For the first circle $x^2+y^2+4x=0$,the center is $C_1 = (-2, 0)$ and the radius is $r_1 = \sqrt{(-2)^2 + 0^2 - 0} = 2$.For the second circle $x^2+y^2-2x=0$,the center is $C_2 = (1, 0)$ and the radius is $r_2 = \sqrt{(1)^2 + 0^2 - 0} = 1$.The distance between the centers is $C_1C_2 = \sqrt{(1 - (-2))^2 + (0 - 0)^2} = \sqrt{3^2} = 3$.Since $r_1 + r_2 = 2 + 1 = 3$,we have $C_1C_2 = r_1 + r_2$.This condition implies that the two circles touch each other externally.When two circles touch each other externally,the number of common tangents is $3$ (two direct common tangents and one transverse common tangent).

The number of common tangents to the circles $x^2+y^2+4x=0$ and $x^2+y^2-2x=0$ is

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