A circle's center is at #(9 ,1 )# and it passes through #(5 ,4 )#. What is the length of an arc covering #(pi ) /3 # radians on the circle?

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Answer 1 · 2018-06-30T10:24:07

Equation circle centered in a generic point is

#(x-x_0)^2+(y-y_0)^2=r^2#

In our case #(x-9)^2+(y-1)^2=r^2# where #r# is unknown

Circle pases by #(5,4)#, then

#(5-9)^2+(4-1)^2=r^2# or

#16+9=25=5^2#

So the radius is #5#

Now: we know that a circle has a lenght of #2pir# in this case the total lenth is #10pi#. The arc lenght for #pi/3# rads will be proportional:

If #2pi# rads has a lenght of #10pi#, then #pi/3# will have?

#cancelpi/3·(10pi)/(2cancelpi)=10/6pi=5/3pi#

Answer 2 · 2018-06-30T10:47:45

The length of the arc is : #l=(5pi)/3~~5.24#

We have a circle with center #C(9,1)# and it passes through #P(5,4).#

Let , #r# be the radius of circle .

Let , #l# be the length of an arc covering # (pi/3)^R # on the circle.

#i.e. theta=pi/3#

We know that the radius of circle is

#r=CP#

#=>r^2=(CP)^2#

#=>r^2=(9-5)^2+(1-4)^2#

#=r^2=16+9=25#

#=>color(blue)(r=5# #andcolor(blue)( theta=pi/3#

So, the length of the arc is :

#color(red)(l=r*theta)=5*pi/3#

#l=(5pi)/3~~5.24#