A circle's center is at $(2, 1)$ $(2 ,1 )$ and it passes through $(3, 2)$ $(3 ,2 )$ . What is the length of an arc covering $\frac{7 π}{12}$ $(7pi ) /12$ radians on the circle?

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Answer 1 · 2018-04-12T04:27:17

arc length $= \frac{11 \sqrt{2}}{6}$ $=(11sqrt(2))/6$ units

Let A $(2, 1)$ $(2,1)$ be the center of the circle and $B (3, 2)$ $B(3,2)$ be the arc length.

Now, radius of circle $= A B = \sqrt{{(3 - 2)}^{2} + {(2 - 1)}^{2}} = \sqrt{2}$ $=AB=sqrt((3-2)^2+(2-1)^2)=sqrt(2)$

$\to θ =$ $rarrtheta=$ arc length/(radius

$\to = \frac{7 π}{12} = \frac{l}{\sqrt{2}}$ $rarr=(7pi)/12=l/sqrt2$

$\to 7 \cdot (\frac{22}{7}) \cdot \frac{1}{12} = \frac{l}{\sqrt{2}}$ $rarr7*(22/7)*1/12=l/sqrt(2)$

$\to l = \frac{11 \sqrt{2}}{6}$ $rarrl=(11sqrt2)/6$ units

A circle's center is at (2,1)(2 ,1 ) and it passes through (3,2)(3 ,2 ). What is the length of an arc covering 7π12(7pi ) /12 radians on the circle?