A circle's center is at $(2, 1)$ $(2 ,1 )$ and it passes through $(6, 1)$ $(6 ,1 )$ . What is the length of an arc covering $\frac{13 π}{12}$ $(13pi ) /12$ radians on the circle?

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A circle's center is at $(2, 1)$ $(2 ,1 )$ and it passes through $(6, 1)$ $(6 ,1 )$ . What is the length of an arc covering $\frac{13 π}{12}$ $(13pi ) /12$ radians on the circle?

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Answer 1 · 2016-09-21T11:26:32

$arc length = \frac{13 π}{3}$ $color(green)("arc length "= (13pi)/3)$

Explanation:

Let point 1 be $P_{1} \to (x_{1}, y_{1}) = (2, 1)$ $P_1->(x_1,y_1)=(2,1)$

Let point 2 be $P_{2} \to (x_{2}, y_{2}) = (6, 1)$ $P_2->(x_2,y_2)=(6,1)$

Let $θ = \frac{13 π}{12}$ $theta=(13pi)/12$

Let the radius be $r$ $r$

Then the arc length is $r θ$ $rtheta$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$Determine the value of r$ $color(blue)("Determine the value of "r)$

Using the fact that the line from $P_{1} to P_{2}$ $P_1" to "P_2$ can be considered as the hypotenuse of a triangle we apply Pythagoras

$\to {(x_{2} - x_{1})}_{2}^{2} + {(y_{2} - y_{1})}_{2}^{2} = r^{2}$ $->(x_2-x_1)^2+(y_2-y_1)^2=r^2$ .

Thus:

$r = \sqrt{{(x_{2} - x_{1})}_{2}^{2} + {(y_{2} - y_{1})}_{2}^{2}}$ $r=sqrt((x_2-x_1)^2+(y_2-y_1)^2)$

$r = \sqrt{{(6 - 2)}^{2} + {(1 - 1)}^{2}} = \sqrt{4^{2}} = 4$ $r=sqrt((6-2)^2+(1-1)^2) =sqrt(4^2) = 4$
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$Determine arc length$ $color(blue)("Determine arc length")$

$"arc length is "rtheta -> cancel(4)^1xx(13pi)/(cancel(12)^3) =(13pi)/3$

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A circle's center is at $(2, 1)$ $(2 ,1 )$ and it passes through $(6, 1)$ $(6 ,1 )$ . What is the length of an arc covering $\frac{13 π}{12}$ $(13pi ) /12$ radians on the circle?

1 Answer

Explanation:

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A circle's center is at (2 ,1 )(2,1)(2 ,1 ) and it passes through (6 ,1 )(6,1)(6 ,1 ). What is the length of an arc covering (13pi ) /12 13π12(13pi ) /12 radians on the circle?

1 Answer

Explanation:

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A circle's center is at $(2, 1)$ $(2 ,1 )$ and it passes through $(6, 1)$ $(6 ,1 )$ . What is the length of an arc covering $\frac{13 π}{12}$ $(13pi ) /12$ radians on the circle?