The number of common tangents to the circles | vedclass

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Question

(C) For the first circle $x^2 + y^2 - 2x + 4y - 4 = 0$,the center $C_1 = (1, -2)$ and radius $r_1 = \sqrt{1^2 + (-2)^2 - (-4)} = \sqrt{1 + 4 + 4} = 3$

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) For the first circle $x^2 + y^2 - 2x + 4y - 4 = 0$,the center $C_1 = (1, -2)$ and radius $r_1 = \sqrt{1^2 + (-2)^2 - (-4)} = \sqrt{1 + 4 + 4} = 3$.For the second circle $x^2 + y^2 - 8x - 4y + 16 = 0$,the center $C_2 = (4, 2)$ and radius $r_2 = \sqrt{4^2 + 2^2 - 16} = \sqrt{16 + 4 - 16} = 2$.The distance between the centers $C_1$ and $C_2$ is $d = \sqrt{(4 - 1)^2 + (2 - (-2))^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5$.Since $d = r_1 + r_2$ $(5 = 3 + 2)$,the two circles touch each other externally.When two circles touch each other externally,the number of common tangents is $3$.

The number of common tangents to the circles $x^2 + y^2 - 2x + 4y - 4 = 0$ and $x^2 + y^2 - 8x - 4y + 16 = 0$ is:

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