Question

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?

Answer 1

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

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

Explanation:

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`