Points #(2 ,4 )# and #(4 ,1 )# are #(3 pi)/4 # radians apart on a circle. What is the shortest arc length between the points?

Let the radius of the circle be #R#. The length of the shorter arc between #(2,4)# and #(4,1)# would simply be #R({3pi}/4)#. Therefore, the key to solving this problem, and many other geometry problems is to find #R#.

We denote the center of the circle as #O#.

The line segment from #O# to any point on the circle has length #R#. This means that the line segment from #O# to #(2,4)# and the line segment from #O# to #(4,1)# both have length #R#.

Consider the triangle formed by #O#, #(2,4)# and #(4,1)#. Using the Law of Cosines, we can find show that the length of the line segment from #(2,4)# to #(4,1)# is

#sqrt{R^2+R^2-2(R)(R)cos({3pi}/4)} #

#= sqrt2 R sqrt{1-cos({3pi}/4)}#

#= sqrt2 R sqrt{1-sqrt2/2}#

#= sqrt2 R sqrt{2-sqrt2}/sqrt2#

#= sqrt{2-sqrt2}R#

We can also use Pythagorean Theorem to find the exact length between #(2,4)# and #(4,1)#.

#sqrt((4-2)^2 + (1-4)^2) = sqrt13#

To find #R#, we just need to equate the 2 independent expressions for the length of the line segment from #(2,4)# to #(4,1)#.

#sqrt{2-sqrt2}R = sqrt13#

Solving the above equation gives

#R = sqrt{frac{13}{2-sqrt2}}#

# = sqrt{frac{13(2+sqrt2)}{(2-sqrt2)(2+sqrt2)}}#

# = sqrt{frac{13(2+sqrt2)}{2}}#

The arc length is given by

#R((3pi)/4) = sqrt{frac{13(2+sqrt2)}{2}}((3pi)/4)#