A circle has a center at #(3 ,1 )# and passes through #(2 ,1 )#. What is the length of an arc covering #pi/8# radians on the circle?

Question

A circle has a center at #(3   ,1   )# and passes through #(2     ,1    )#.  What is the length of an arc covering #pi/8# radians on the circle?

Violet Aldrich · Accepted Answer

length of arc #s=r*theta=1*pi/8=pi/8=0.392699#

From the given data: A circle has a center at (3,1) and passes through (2,1). What is the length of an arc covering π/8 radians on the circle?

We need the radius r to compute for the arc

#r=sqrt((3-2)^2+(1-1)^2)=sqrt(1)=1#

#s=r*theta=1*pi/8=pi/8=0.392699#

God bless...I hope the explanation is useful.

Sophia Alfred · Answer

undefined To find the length of an arc covering $ \frac{\pi}{8} $ radians on a circle, you can use the formula:

$$ \text{Arc Length} = r \times \theta $$

where $ r $ is the radius of the circle and $ \theta $ is the angle in radians.

Given that the circle has a center at $ (3, 1) $ and passes through $ (2, 1) $, we can find the radius $ r $ using the distance formula:

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

$$ r = \sqrt{(2 - 3)^2 + (1 - 1)^2} $$

$$ r = \sqrt{1 + 0} $$

$$ r = 1 $$

Now, plug in the values of $ r = 1 $ and $ \theta = \frac{\pi}{8} $ into the arc length formula:

$$ \text{Arc Length} = 1 \times \frac{\pi}{8} $$

$$ \text{Arc Length} = \frac{\pi}{8} $$