A circle's center is at #(2 ,4 )# and it passes through #(7 ,6 )#. What is the length of an arc covering #(15pi ) /8 # radians on the circle?

Question

A circle's center is at #(2 ,4 )# and it passes through #(7    ,6   )#.  What is the length of an arc covering #(15pi ) /8  # radians on the circle?

Sadie Allen · Accepted Answer

Length of the arc: #31.7# (approx.)

If the circle has a center at #(2,4)# and passes through #(7,6)# then it has a radius of #color(white)("XXX")r=sqrt((7-2)^2+(6-4)^2)=sqrt(25+4) =sqrt(29)# and a diameter of #color(white)("XXX")d=2r=2sqrt(29)# The circumference of the circle would be #color(white)("XXX")"Circumference"_circ=pid = 2sqrt(29)pi# The complete circle's circumference is covered by an arc of #2pi=(16pi)/8# An arc of #(15pi)/8# will cover #((15pi)/8)/((16pi)/8)=15/16# of #"Circumference"_circ# The given arc will cover #color(white)("XXX")15/16xx2sqrt(29)pi# #color(white)("XXX")~~31.72123912# (with a calculator)

Carson Arnold · Answer

undefined To find the length of an arc covering $ \frac{15\pi}{8} $ radians on the circle, we need to first find the radius of the circle using the given points.

Given:
- Center of the circle: $ (2, 4) $
- Point on the circle: $ (7, 6) $

Using the distance formula between two points, we can find the radius:

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

Substituting the given points:

$$ \text{Distance} = \sqrt{(7 - 2)^2 + (6 - 4)^2} $$

$$ = \sqrt{5^2 + 2^2} $$

$$ = \sqrt{25 + 4} $$

$$ = \sqrt{29} $$

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

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

Given:
- $ \text{Angle} = \frac{15\pi}{8} $
- $ r = \sqrt{29} $

$$ \text{Length of Arc} = \sqrt{29} \times \frac{15\pi}{8} $$

$$ = \frac{15\pi \sqrt{29}}{8} $$

So, the length of an arc covering $ \frac{15\pi}{8} $ radians on the circle is $ \frac{15\pi \sqrt{29}}{8} $.