A circle has a center at #(7 ,9 )# and passes through #(1 ,1 )#. What is the length of an arc covering #(3pi ) /4 # radians on the circle?

Question

A circle has a center at #(7  ,9  )# and passes through #(1     ,1    )#.  What is the length of an arc covering #(3pi ) /4  # radians on the circle?

Aurora Ayala · Accepted Answer

#(15pi)/2# First thing to do is to find the length of the radius. The line segment joining the center of the circle to any point on the circle constitutes a radius. Therefore, the line segment joining #(7,9)# and #(1,1)# is a radius. To find its length, you can use Pythagoras Theorem. #r = sqrt{(7 - 1)^2 + (9 - 1)^2} = 10# Next, you should know that an arc subtending an angle of #theta# in radians, has arc length, s, given by #s = r theta#. #s = (10)*((3pi)/4) = (15pi)/2#

Sadie Allen · Answer

undefined To find the length of an arc covering $ \frac{3\pi}{4} $ radians on the circle, we need to first find the radius of the circle using the given center and one point on the circle. Then, we can use the formula for the arc length of a circle to calculate the length of the arc.

The formula for the radius ($ r $) of a circle given its center ($ (h, k) $) and a point ($ (x_1, y_1) $) on the circle is:

$$ r = \sqrt{(x_1 - h)^2 + (y_1 - k)^2} $$

Given the center $ (h, k) = (7, 9) $ and a point $ (x_1, y_1) = (1, 1) $, we can calculate the radius ($ r $):

$$ r = \sqrt{(1 - 7)^2 + (1 - 9)^2} $$

$$ r = \sqrt{(-6)^2 + (-8)^2} $$

$$ r = \sqrt{36 + 64} $$

$$ r = \sqrt{100} $$

$$ r = 10 $$

Now that we have the radius ($ r = 10 $), we can use the formula for the arc length of a circle:

$$ \text{Arc Length} = r \times \text{angle in radians} $$

$$ \text{Arc Length} = 10 \times \frac{3\pi}{4} $$

$$ \text{Arc Length} = \frac{30\pi}{4} $$

$$ \text{Arc Length} = \frac{15\pi}{2} $$

So, the length of the arc covering $ \frac{3\pi}{4} $ radians on the circle is $ \frac{15\pi}{2} $ units.