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.