A circle's center is at $(9, 7)$ $(9 ,7 )$ and it passes through $(6, 2)$ $(6 ,2 )$ . What is the length of an arc covering $\frac{5 π}{6}$ $(5 pi ) /6$ radians on the circle?

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A circle's center is at $(9, 7)$ $(9 ,7 )$ and it passes through $(6, 2)$ $(6 ,2 )$ . What is the length of an arc covering $\frac{5 π}{6}$ $(5 pi ) /6$ radians on the circle?

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Answer 1 · 2016-11-02T18:27:16

First of all the radius of the circle would be the distance between the center at (9,7) and the point (6,2) on the circle. This would be $\sqrt{{(9 - 6)}^{2} + {(7 - 2)}^{2}}$ $sqrt((9-6)^2 + (7-2)^2)$ = $\sqrt{34}$ $sqrt34$ cms

Arc length formula is $s = r θ$ $s= rtheta$ where angle $(θ)$ $(theta)$ is measured in radians. In the present case $θ = \frac{5 π}{6}$ $theta = (5pi)/6$ , hence arc length would be $\sqrt{34} \cdot \frac{5 π}{6}$ $sqrt34 * (5pi)/6$ = 15.26 cms

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A circle's center is at $(9, 7)$ $(9 ,7 )$ and it passes through $(6, 2)$ $(6 ,2 )$ . What is the length of an arc covering $\frac{5 π}{6}$ $(5 pi ) /6$ radians on the circle?

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A circle's center is at (9,7)(9 ,7 ) and it passes through (6,2)(6 ,2 ). What is the length of an arc covering 5π6(5 pi ) /6 radians on the circle?

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A circle's center is at $(9, 7)$ $(9 ,7 )$ and it passes through $(6, 2)$ $(6 ,2 )$ . What is the length of an arc covering $\frac{5 π}{6}$ $(5 pi ) /6$ radians on the circle?