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 $0.6008pi$ .

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