Points #(2 ,8 )# and #(1 ,5 )# are #( pi)/4 # radians apart on a circle. What is the shortest arc length between the points?

1 Answer
May 15, 2016

The arc length is
#L=pi sqrt(5(2+sqrt(2)))/4#

Explanation:

We have coordinates of two points:
Point #A(2,8)#
Point #B(1,5)#

Consider point #O# with unknown coordinates to be a center of a circle that goes through points #A# and #B#.
Then we know that angle #/_AOB = pi/4#.

To define the length of an arc from #A# to #B# we need to know the radius of this circle #R=OA=OB#. Then, knowing the central angle #/_AOB=pi/4#, we can easily determine the length of an arc from #A# to #B# as #(piR)/4#.

To determine the radius #R=OA=OB#, draw a perpendicular bisector to a segment #AB#. It should pass through a center of a circle #O#.

Let the midpoint of a segment #AB# be point #M#.
Consider triangle #Delta AOM#.
It's a right triangle with hypotenuse #AO# (that is, the radius #R# of a circle), cathetus #AM# being equal to half of a distance from #A# to #B# and an opposite acute angle #/_AOM=pi/8#.

#AM=1/2 AB = 1/2 sqrt((2-1)^2+(8-5)^2)=sqrt(10)/2#

Now we can determine
#R=AO=(AM)/sin(/_AOM)=sqrt(10)/(2sin(pi/8))#

To determine exact value of #sin(pi/8)#, notice that
#sin^2(pi/4)=[2sin(pi/8)Cos(pi/8)]^2=4sin^2(pi/8)[1-sin^2(pi/8)]#

Let #sin^2(pi/8)=x#.

Then, since #sin^2(pi/4)=(sqrt(2)/2)^2=1/2#,
#4x(1-x)=1/2#
or
#8x^2-8x+1=0#

Choosing a smaller root of this quadratic equation,
#x=(2-sqrt(2))/4#

Therefore,
#sin(pi/8) = 1/2sqrt(2-sqrt(2))#

Hence,
#R=sqrt(10)/sqrt(2-sqrt(2))=sqrt(10/(2-sqrt(2))=#
#=sqrt(10(2+sqrt(2))/[(2-sqrt(2))(2+sqrt(2))])=#
#=sqrt(10(2+sqrt(2))/2)=sqrt(5(2+sqrt(2))#

The arc length is
#L=piR/4=pi sqrt(5(2+sqrt(2)))/4#