The number of direct common tangents to the c | vedclass

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(C) For the first circle $x^2 + y^2 = 4$,the center $C_1 = (0, 0)$ and radius $r_1 = \sqrt{4} = 2$.For the second circle $x^2 + y^2 - 8x - 8y + 7 = 0$

Vedclass · Accepted Answer

$2$

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(C) For the first circle $x^2 + y^2 = 4$,the center $C_1 = (0, 0)$ and radius $r_1 = \sqrt{4} = 2$.For the second circle $x^2 + y^2 - 8x - 8y + 7 = 0$,the center $C_2 = (-(-8/2), -(-8/2)) = (4, 4)$ and radius $r_2 = \sqrt{4^2 + 4^2 - 7} = \sqrt{16 + 16 - 7} = \sqrt{25} = 5$.The distance between the centers $C_1 C_2 = \sqrt{(4-0)^2 + (4-0)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} \approx 5.66$.Since $|r_1 - r_2| When two circles intersect,they have $2$ direct common tangents and $0$ transverse common tangents.Therefore,the number of direct common tangents is $2$.

The number of direct common tangents to the circles $x^2 + y^2 = 4$ and $x^2 + y^2 - 8x - 8y + 7 = 0$ is:

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