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 $\frac{2 π}{3}$ $(2pi ) /3$ radians on the circle?

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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 $\frac{2 π}{3}$ $(2pi ) /3$ radians on the circle?

1 Answer

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Answer 1 · 2018-03-18T03:17:01

Arc Length $= r \cdot θ \approx 4.68$ $= r * theta ~~ 4.68$

Explanation:

$C e n t r e (2, 6), passing thru (3, 8), θ = \frac{2 π}{3}$ $Centre (2,6), "passing thru" (3,8), theta = (2pi)/3$

Hence $Radius r = \sqrt{{(3 - 2)}^{2} + {(8 - 6)}^{2}} = \sqrt{5}$ $"Radius " r = sqrt((3-2)^2 + (8-6)^2) = sqrt5$ , using distance formula

Arc Length $= r \cdot θ = \sqrt{5} \cdot (\frac{2 π}{3}) = (\frac{2}{3}) \sqrt{5} π \approx 4.68$ $= r * theta = sqrt5 * ((2pi)/3) = (2/3) sqrt5 pi ~~ 4.68$

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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 $\frac{2 π}{3}$ $(2pi ) /3$ radians on the circle?

1 Answer

Explanation:

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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 $\frac{2 π}{3}$ $(2pi ) /3$ radians on the circle?