Points #(3 ,2 )# and #(8 ,1 )# are #(2 pi)/3 # radians apart on a circle. What is the shortest arc length between the points?

From the given data: Points (3,2) and (8,1) are 2π/3 radians apart on a circle. What is the shortest arc length between the points?

This means , we have the central angle #theta=(2pi)/3#. To determine the arc #s#, we need to know the radius #r#.

Let us solve #r#:
Let #(x, y)# be the center of the circle

#r=r#

#(3-x)^2+(2-y)^2=(8-x)^2+(1-y)^2#
After simplifying this equation, we have

#color(red)(5x-y-26=0" ")##color(blue)("first equation")#

Making use of the slopes of the radii:

Let #m_2=(y-1)/(x-8)# and #m_1=(y-2)/(x-3)#

#Tan theta=Tan ((2pi)/3)=(m_2-m_1)/(1+m_2*m_1)#

#Tan((2pi)/3)=-sqrt3=((y-1)/(x-8)-(y-2)/(x-3))/(1+(y-1)/(x-8)*(y-2)/(x-3))#

After simplifying this equation

#color(red)(x+5y-13=-sqrt3*(x^2+y^2-11x-3y+26))##color(blue)("second equation")#

Use now, the first and second equations to solve for the center (x, y)

We have

#color(red)(26sqrt3x^2+(26-286sqrt3)x+780sqrt3-143=0)#

By Quadratic Equation Formula

#x=5.78868# and #y=2.9434#

Compute now for radius #r# using center (x, y) and point on the circle (8, 1)

#r=sqrt((8-5.78868)^2+(1-2.9434)^2)#

#r=2.94393" "#units

Compute now for the arc

#s=r*theta=2.94393*(2pi)/3=6.16576" "#units

God bless....I hope the explanation is useful.