A circle's center is at $(4 ,0 )$ and it passes through $(6 ,9 )$ . What is the length of an arc covering $(5pi ) /3$ radians on the circle?

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Answer 1 · 2016-09-29T18:49:58

$s = sqrt(85)*(5pi)/3 ~~ 48.27$

First we must find the radius of the circle.

We know the distance from the center to any point it passes through is the radius.

So, using the distance formula: $d = sqrt((x_2-x_1)^2+(y_2-y_1)^2)$

$d = sqrt((4-6)^2+(0-9)^2) = sqrt(85)$

So, the radius is $sqrt(85)$

And we know the formula for arc length is: $s = rtheta$
Where s is the arc length, r is the radius, and theta is the angle in radians.

Plugging in, $s = sqrt(85)*(5pi)/3$

Answer 2 · 2016-09-29T18:51:16

$s = 5sqrt(85)pi/3$

Let r = the radius

$r = sqrt((6 - 4)^2 + (9 - 0)^2)$

$r = sqrt(85)$

Let s = the arc length

$s = rtheta$ where $theta$ is the given angle

$s = 5sqrt(85)pi/3$

A circle's center is at (4 ,0 )(4 ,0 ) and it passes through (6 ,9 )(6 ,9 ). What is the length of an arc covering (5pi ) /3 (5pi ) /3 radians on the circle?