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 $\frac{5 π}{3}$ $(5pi ) /3$ radians on the circle?

Question

1861 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-29T18:49:58

$s = \sqrt{85} \cdot \frac{5 π}{3} \approx 48.27$ $s = sqrt(85)*(5pi)/3 ~~ 48.27$

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: $d = \sqrt{{(x_{2} - x_{1})}_{2}^{2} + {(y_{2} - y_{1})}_{2}^{2}}$ $d = sqrt((x_2-x_1)^2+(y_2-y_1)^2)$

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

So, the radius is $\sqrt{85}$ $sqrt(85)$

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

Plugging in, $s = \sqrt{85} \cdot \frac{5 π}{3}$ $s = sqrt(85)*(5pi)/3$

Answer 2 · 2016-09-29T18:51:16

$s = 5 \sqrt{85} \frac{π}{3}$ $s = 5sqrt(85)pi/3$

Let r = the radius

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

$r = \sqrt{85}$ $r = sqrt(85)$

Let s = the arc length

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

$s = 5 \sqrt{85} \frac{π}{3}$ $s = 5sqrt(85)pi/3$

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 5π3(5pi ) /3 radians on the circle?