A circle's center is at $(2 ,6 )$ and it passes through $(3 ,8 )$ . What is the length of an arc covering $(2pi ) /3$ radians on the circle?

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Answer 1 · 2018-03-18T03:17:01

Arc Length $= r * theta ~~ 4.68$

$Centre (2,6), "passing thru" (3,8), theta = (2pi)/3$

Hence $"Radius " r = sqrt((3-2)^2 + (8-6)^2) = sqrt5$ , using distance formula

Arc Length $= r * theta = sqrt5 * ((2pi)/3) = (2/3) sqrt5 pi ~~ 4.68$

A circle's center is at (2 ,6 )(2 ,6 ) and it passes through (3 ,8 )(3 ,8 ). What is the length of an arc covering (2pi ) /3 (2pi ) /3 radians on the circle?