A circle's center is at #(9 ,4 )# and it passes through #(1 ,6 )#. What is the length of an arc covering #(2pi ) /3 # radians on the circle?

Question

A circle's center is at #(9 ,4 )# and it passes through #(1    ,6    )#.  What is the length of an arc covering #(2pi ) /3  # radians on the circle?

Savannah Aguilar · Accepted Answer

#=17.2# unit

radius of the circle #r=sqrt((9-1)^2+(4-6)^2)=sqrt68 # angle subtended buy arc at center# (theta=2pi/3)# Arc langth #S=rxxtheta=sqrt68xx2xxpi/3=17.2#

Parker Allen · Answer

undefined To find the length of an arc covering $ \frac{2\pi}{3} $ radians on the circle, we need to calculate the radius of the circle first using the given coordinates of the center and a point on the circle. Then, we can use the formula for the arc length of a circle, which is $ s = r \times \theta $, where $ s $ is the arc length, $ r $ is the radius of the circle, and $ \theta $ is the angle in radians subtended by the arc.

Given:
Center of the circle: $ (9, 4) $
Point on the circle: $ (1, 6) $
Angle $ \theta = \frac{2\pi}{3} $

Using the distance formula to find the radius:
$$ r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$

Substituting the coordinates of the center and the point on the circle:
$$ r = \sqrt{(1 - 9)^2 + (6 - 4)^2} $$
$$ r = \sqrt{(-8)^2 + (2)^2} $$
$$ r = \sqrt{64 + 4} $$
$$ r = \sqrt{68} $$
$$ r = 2\sqrt{17} $$

Now, we can calculate the arc length using the formula:
$$ s = r \times \theta $$
$$ s = (2\sqrt{17}) \times \frac{2\pi}{3} $$
$$ s = \frac{4\pi\sqrt{17}}{3} $$

Therefore, the length of the arc covering $ \frac{2\pi}{3} $ radians on the circle is $ \frac{4\pi\sqrt{17}}{3} $ units.