# 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?

Jun 30, 2018

Equation circle centered in a generic point is

${\left(x - {x}_{0}\right)}^{2} + {\left(y - {y}_{0}\right)}^{2} = {r}^{2}$

In our case ${\left(x - 9\right)}^{2} + {\left(y - 1\right)}^{2} = {r}^{2}$ where $r$ is unknown

Circle pases by $\left(5 , 4\right)$, then

${\left(5 - 9\right)}^{2} + {\left(4 - 1\right)}^{2} = {r}^{2}$ or

$16 + 9 = 25 = {5}^{2}$

So the radius is $5$

Now: we know that a circle has a lenght of $2 \pi r$ in this case the total lenth is $10 \pi$. The arc lenght for $\frac{\pi}{3}$ rads will be proportional:

If $2 \pi$ rads has a lenght of $10 \pi$, then $\frac{\pi}{3}$ will have?

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

Jun 30, 2018

The length of the arc is : $l = \frac{5 \pi}{3} \approx 5.24$

#### Explanation:

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

Let , $r$ be the radius of circle .

Let , $l$ be the length of an arc covering ${\left(\frac{\pi}{3}\right)}^{R}$ on the circle.

$i . e . \theta = \frac{\pi}{3}$

We know that the radius of circle is

$r = C P$

$\implies {r}^{2} = {\left(C P\right)}^{2}$

$\implies {r}^{2} = {\left(9 - 5\right)}^{2} + {\left(1 - 4\right)}^{2}$

$= {r}^{2} = 16 + 9 = 25$

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

So, the length of the arc is :

$\textcolor{red}{l = r \cdot \theta} = 5 \cdot \frac{\pi}{3}$

$l = \frac{5 \pi}{3} \approx 5.24$