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

Question

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

Violet Aldrich · Accepted Answer

#~= 9.911# #r=sqrt((5-2)^2+(5-7)^2)=sqrt(9+4)=sqrt(13)# so length of arc is: #sqrt(13)xx(7pi)/8 ~= 9.911#

Parker Allen · Answer

undefined To find the length of an arc covering $ \frac{7\pi}{8} $ radians on the circle, you need to use the formula for the length of an arc of a circle. First, find the radius of the circle using the distance formula between the center of the circle and a point on its circumference:

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

Substitute the coordinates of the center $(5, 5)$ and the point $(2, 7)$:

$$ \text{Radius} = \sqrt{(2 - 5)^2 + (7 - 5)^2} = \sqrt{9 + 4} = \sqrt{13} $$

Now, the formula for the length of an arc of a circle is given by:

$$ \text{Arc Length} = \text{Radius} \times \text{Angle in Radians} $$

Substitute the given angle $ \frac{7\pi}{8} $ and the radius $ \sqrt{13} $:

$$ \text{Arc Length} = \sqrt{13} \times \frac{7\pi}{8} $$

Simplify:

$$ \text{Arc Length} = \frac{7\sqrt{13}\pi}{8} $$

So, the length of the arc covering $ \frac{7\pi}{8} $ radians on the circle is $ \frac{7\sqrt{13}\pi}{8} $ units.