A circle's center is at $(2, 5)$ $(2 ,5 )$ and it passes through $(1, 4)$ $(1 ,4 )$ . What is the length of an arc covering $\frac{π}{3}$ $( pi ) /3$ radians on the circle?

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A circle's center is at $(2, 5)$ $(2 ,5 )$ and it passes through $(1, 4)$ $(1 ,4 )$ . What is the length of an arc covering $\frac{π}{3}$ $( pi ) /3$ radians on the circle?

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Answer 1 · 2016-02-17T03:59:01

length of arc $= \frac{\sqrt{2} \cdot π}{3} = 1.48096$ $=(sqrt2*pi)/3=1.48096" "$ units

Explanation:

To solve for the length of arc s:
Use the formula

$s = r \cdot θ$ $s=r*theta$ using the given data: $θ = \frac{π}{3}$ $theta=pi/3$ and computed value of $r$ $r$

to compute for r: use distance formula with points center $(x_{c}, y_{c}) = (2, 5)$ $(x_c, y_c)=(2, 5)$ and the given point $(x_{1}, y_{1}) = (1, 4)$ $(x_1, y_1)=(1, 4)$

$r = \sqrt{{(x_{c} - x_{1})}_{c}^{2} + {(y_{c} - y_{1})}_{c}^{2}}$ $r=sqrt((x_c-x_1)^2+(y_c-y_1)^2)$

$r = \sqrt{{(2 - 1)}^{2} + {(5 - 4)}^{2}}$ $r=sqrt((2-1)^2+(5-4)^2)$

$r = \sqrt{2}$ $r=sqrt2$

Solve arc length $s$ $s$

$s = r \cdot θ = \sqrt{2} \cdot \frac{π}{3}$ $s=r*theta=sqrt2 *pi/3$

$s = \frac{\sqrt{2} \cdot π}{3} = 1.48096$ $s=(sqrt2*pi)/3=1.48096" "$ units

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A circle's center is at $(2, 5)$ $(2 ,5 )$ and it passes through $(1, 4)$ $(1 ,4 )$ . What is the length of an arc covering $\frac{π}{3}$ $( pi ) /3$ radians on the circle?

1 Answer

Explanation:

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A circle's center is at $(2, 5)$ $(2 ,5 )$ and it passes through $(1, 4)$ $(1 ,4 )$ . What is the length of an arc covering $\frac{π}{3}$ $( pi ) /3$ radians on the circle?