A circle's center is at $(5, 9)$ $(5 ,9 )$ and it passes through $(7, 3)$ $(7 ,3 )$ . What is the length of an arc covering $\frac{15 π}{8}$ $(15pi ) /8$ radians on the circle?

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A circle's center is at $(5, 9)$ $(5 ,9 )$ and it passes through $(7, 3)$ $(7 ,3 )$ . What is the length of an arc covering $\frac{15 π}{8}$ $(15pi ) /8$ radians on the circle?

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Answer 1 · 2017-08-27T04:46:49

$\frac{15 \sqrt{10} π}{4} \approx 37.25$ $(15sqrt10pi)/4~~37.25$ units

Explanation:

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Given circle's center $O$ $O$ is at $(5, 9)$ $(5,9)$ and the circle passes through $(7, 3)$ $(7,3)$ ,
$\Rightarrow$ $=>$ radius of the circle $r = \sqrt{{(7 - 5)}^{2} + {(3 - 9)}^{2}} = \sqrt{4 + 36} = \sqrt{40}$ $r=sqrt((7-5)^2+(3-9)^2)=sqrt(4+36)=sqrt40$

Arc length $= r θ$ $=rtheta$ ( $θ$ $theta$ in radians)
given $θ = \frac{15 π}{8}$ $theta=(15pi)/8$ ,
arc length $= \sqrt{40} \times \frac{15 π}{8} = 2 \sqrt{10} \times \frac{15 π}{8}$ $= sqrt40xx(15pi)/8=2sqrt10xx(15pi)/8$
$= \frac{15 \sqrt{10} \cdot π}{4} \approx 37.25$ $=(15sqrt10*pi)/4~~37.25$ units

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A circle's center is at $(5, 9)$ $(5 ,9 )$ and it passes through $(7, 3)$ $(7 ,3 )$ . What is the length of an arc covering $\frac{15 π}{8}$ $(15pi ) /8$ radians on the circle?

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Explanation:

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A circle's center is at (5,9)(5 ,9 ) and it passes through (7,3)(7 ,3 ). What is the length of an arc covering 15π8(15pi ) /8 radians on the circle?

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A circle's center is at $(5, 9)$ $(5 ,9 )$ and it passes through $(7, 3)$ $(7 ,3 )$ . What is the length of an arc covering $\frac{15 π}{8}$ $(15pi ) /8$ radians on the circle?