The number of common tangents to the circles | vedclass

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(C) For the first circle $x^2 + y^2 = 1$,the center $C_1 = (0, 0)$ and radius $r_1 = 1$.For the second circle $x^2 + y^2 - 4x + 3 = 0$,we rewrite it a

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) For the first circle $x^2 + y^2 = 1$,the center $C_1 = (0, 0)$ and radius $r_1 = 1$.For the second circle $x^2 + y^2 - 4x + 3 = 0$,we rewrite it as $(x-2)^2 + y^2 = 1$,so the center $C_2 = (2, 0)$ and radius $r_2 = 1$.The distance between the centers $C_1$ and $C_2$ is $d = \sqrt{(2-0)^2 + (0-0)^2} = 2$.Since $d = r_1 + r_2 = 1 + 1 = 2$,the two circles touch each other externally.When two circles touch externally,the number of common tangents is $3$.

The number of common tangents to the circles $x^2 + y^2 = 1$ and $x^2 + y^2 - 4x + 3 = 0$ is

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