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

1 Answer
Jan 22, 2016

#(5sqrt(29)pi)/6#

Explanation:

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

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

and the arc length with #theta= (5pi)/6# is
#color(white)("XXX")(5pi)/6*sqrt(29) = (5sqrt(29)pi)/6#