A circle's center is at #(2 ,1 )# and it passes through #(6 ,1 )#. What is the length of an arc covering #(13pi ) /12 # radians on the circle?

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Answer 1 · 2016-09-21T11:26:32

#color(green)("arc length "= (13pi)/3)#

Let point 1 be #P_1->(x_1,y_1)=(2,1)#

Let point 2 be #P_2->(x_2,y_2)=(6,1)#

Let #theta=(13pi)/12#

Let the radius be #r#

Then the arc length is #rtheta#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine the value of "r)#

Using the fact that the line from #P_1" to "P_2# can be considered as the hypotenuse of a triangle we apply Pythagoras

#->(x_2-x_1)^2+(y_2-y_1)^2=r^2#.

Thus:

#r=sqrt((x_2-x_1)^2+(y_2-y_1)^2)#

#r=sqrt((6-2)^2+(1-1)^2) =sqrt(4^2) = 4#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine arc length")#

#"arc length is "rtheta -> cancel(4)^1xx(13pi)/(cancel(12)^3) =(13pi)/3#