A circle's center is at #(8 ,7 )# and it passes through #(6 ,2 )#. What is the length of an arc covering #(5 pi ) /6 # radians on the circle?

Answer 1

#(5sqrt(29)pi)/6#

The length of an arc with radius #r# covering #theta# radians is #color(white)("XXX")theta*r# (Think of this in terms of a complete circle whose circumference is #2pir#).
If the arc has a center at #(8,7)# and passes through #(6,2)# then its radius is #color(white)("XXX")r=sqrt((8-6)^2+(7-2)^2) = sqrt(2^2+5^2) = sqrt(29)#
and the arc length with #theta= (5pi)/6# is #color(white)("XXX")(5pi)/6*sqrt(29) = (5sqrt(29)pi)/6#
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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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