A circle's center is at #(3 ,4 )# and it passes through #(0 ,2 )#. What is the length of an arc covering #( pi ) /6 # radians on the circle?

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Answer 1 · 2016-01-22T23:20:35

Center of circle is at #(3,4)#, Circle passes through #(0,2)#
Angle made by arc on the circle=#pi/6#, Length of arc# =??#

Let #C=(3,4)#, #P=(0,2)#

Calculating distance between #C# and #P# will giveus the radius of the circle.

#|CP|=sqrt((0-3)^2+(2-4)^2)=sqrt(9+4)=sqrt13#

Let the radius be denoted by #r#, the angle subtended by the arc at the center be denoted by #theta# and the length of the arc be denoted by #s#.

Then #r=sqrt13# and #theta=pi/6#

We know that:
#s=rtheta#

#implies s=sqrt13*pi/6=3.605/6*pi=0.6008pi#

#implies s=0.6008pi#

Hence, the length of arc is #0.6008pi#.