# A circle has a center at (3 ,5 ) and passes through (0 ,2 ). What is the length of an arc covering (3pi ) /4  radians on the circle?

Mar 10, 2016

${l}_{a} = \frac{9}{4} \sqrt{2} \pi$.

#### Explanation:

The radius of the circle with center in $C \left(3 , 5\right)$ that passes through $A \left(0 , 2\right)$ is the distance between the two points:

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

$= \sqrt{{\left(0 - 3\right)}^{2} + {\left(2 - 5\right)}^{2}} = \sqrt{9 + 9} = \sqrt{2 \cdot 9} = 3 \sqrt{2}$.

The angle at the center of a circle is, obviously, $2 \pi$, so the angle of $\frac{3}{4} \pi$ is: $\frac{\frac{3}{4} \pi}{2 \pi} = \frac{3}{8}$ of the whole circle.

Since the lenght of a circle is: ${l}_{c} = 2 \pi r = 2 \pi \cdot 3 \sqrt{2} = 6 \sqrt{2} \pi$, then the lenght of the arc is:

${l}_{a} = 6 \sqrt{2} \pi \cdot \frac{3}{8} = \frac{9}{4} \sqrt{2} \pi$.