A circle's center is at $(7 ,2 )$ and it passes through $(5 ,6 )$ . What is the length of an arc covering $(7pi ) /4$ radians on the circle?

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Answer 1 · 2016-09-25T09:02:59

$:. "Length of Arc="(7sqrt5pi)/2~~24.59"unit"$ .

Let $r$ be the radius of the circle.

The Centre C of the circle is $C(7,2)$ and, the circle passes through

the pt. $P(5,6)$ . Hence, the dist. $CP=r$ .

Using the Distance Formula, we get,

$r^2=CP^2=(7-5)^2+(2-6)^2=4+16=20$

$:. r=sqrt20=2sqrt5$ .

Hence, the $"Length of Arc CP=s="rtheta$ , where, $theta$ is the

measure (in Radians ), of the $/_$ made by the $"Arc "CP$ at the

Centre. We have, $r=2sqrt5, &, theta=(7pi)/4.$

$:. "Length of Arc="2sqrt5*(7pi)/4=(7sqrt5pi)/2~~24.59"unit"$ .

A circle's center is at (7 ,2 )(7 ,2 ) and it passes through (5 ,6 )(5 ,6 ). What is the length of an arc covering (7pi ) /4 (7pi ) /4 radians on the circle?