A circle's center is at $(8, 7)$ $(8 ,7 )$ and it passes through $(6, 2)$ $(6 ,2 )$ . What is the length of an arc covering $\frac{5 π}{6}$ $(5 pi ) /6$ radians on the circle?

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A circle's center is at $(8, 7)$ $(8 ,7 )$ and it passes through $(6, 2)$ $(6 ,2 )$ . What is the length of an arc covering $\frac{5 π}{6}$ $(5 pi ) /6$ radians on the circle?

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Answer 1 · 2016-01-22T17:11:25

$\frac{5 \sqrt{29} π}{6}$ $(5sqrt(29)pi)/6$

Explanation:

The length of an arc with radius $r$ $r$ covering $θ$ $theta$ radians is
$XXX θ \cdot r$ $color(white)("XXX")theta*r$
(Think of this in terms of a complete circle whose circumference is $2 π r$ $2pir$ ).

If the arc has a center at $(8, 7)$ $(8,7)$ and passes through $(6, 2)$ $(6,2)$ then its radius is
$XXX r = \sqrt{{(8 - 6)}^{2} + {(7 - 2)}^{2}} = \sqrt{2^{2} + 5^{2}} = \sqrt{29}$ $color(white)("XXX")r=sqrt((8-6)^2+(7-2)^2) = sqrt(2^2+5^2) = sqrt(29)$

and the arc length with $θ = \frac{5 π}{6}$ $theta= (5pi)/6$ is
$XXX \frac{5 π}{6} \cdot \sqrt{29} = \frac{5 \sqrt{29} π}{6}$ $color(white)("XXX")(5pi)/6*sqrt(29) = (5sqrt(29)pi)/6$

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A circle's center is at $(8, 7)$ $(8 ,7 )$ and it passes through $(6, 2)$ $(6 ,2 )$ . What is the length of an arc covering $\frac{5 π}{6}$ $(5 pi ) /6$ radians on the circle?

1 Answer

Explanation:

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A circle's center is at $(8, 7)$ $(8 ,7 )$ and it passes through $(6, 2)$ $(6 ,2 )$ . What is the length of an arc covering $\frac{5 π}{6}$ $(5 pi ) /6$ radians on the circle?