Points #(7 ,1 )# and #(8 ,9 )# are #(3 pi)/4 # radians apart on a circle. What is the shortest arc length between the points?

1 Answer
Jul 19, 2016

#color(blue)((3sqrt(65))/(4sqrt(2+sqrt(2)))#

(This is pretty ugly; could someone check this please?)

Explanation:

Distance between #(7,1)# and #(8,9)# is given by the Pythagorean Theorem as
#color(white)("XXX")sqrt(1^2+8^2)=sqrt(65)#

To determine the arc length we will need to determine the radius of a circle with the given points at an angle of #(3pi)/4# relative to the center of the circle.
enter image source here
(Note this diagram is not accurate).

If we let the radius of this circle be #color(green)(r)#
and denote
#color(white)("XXX")(7,1)# as #P#,
#color(white)("XXX")(8,9)# as #Q#,
#color(white)("XXX")#the center of the circle as #C#, and
#color(white)("XXX")#the point on the extension of #PC# to form a right angle with #Q# as #S#

then
Since
#color(white)("XXX")/_PSQ=pi/2#, and
#color(white)("XXX")/_QCS=pi-(3pi)/4=pi/4#
#rarr#
#color(white)("XXX")/_CQS=pi/4# and
#color(white)("XXX")abs(CS)=abs(QS)=r/sqrt(2)#

Therefore
#color(white)("XXX")abs(PS)=r+r/sqrt(2)=((sqrt(2)+1)/sqrt(2))r#
and
#color(white)("XXX")abs(PQ)=sqrt((r/sqrt(2))^2+(((sqrt(2)+1)/(sqrt(2)))r)^2)#
#color(white)("XXX")=(sqrt(2+sqrt(2)))r#

But we previously determined that
#color(white)("XXX")abs(PQ)=sqrt(65)#
So
#color(white)("XXX")r=sqrt(65)/(sqrt(2+sqrt(2))#

The shortest arc length of an arc with radius #color(green)(r)# and an angle of #(3pi)/4# is #3/4r#

#rArr # shortest arc length is #3/4xxsqrt(65)/(sqrt(2+sqrt(2)))#

#color(white)("XXX")=(3sqrt(65))/(4sqrt(2+sqrt(2)))#