# A circle has a center at (7 ,6 ) and passes through (2 ,1 ). What is the length of an arc covering pi/8 radians on the circle?

Jan 27, 2016

$\frac{5 \sqrt{2} \pi}{8}$

#### Explanation:

We know the center and a point on the circle. The distance between the two points is the radius (draw a picture and convince yourself this).

We know that the distance between two points $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$ on the Euclidean plane is given by $d = \sqrt{{\left({x}_{1} - {x}_{2}\right)}^{2} + {\left({y}_{1} - {y}_{2}\right)}^{2}}$.
Thus, radius $r = \sqrt{{\left(7 - 2\right)}^{2} + {\left(6 - 1\right)}^{2}} = \sqrt{{5}^{2} + {5}^{2}} = \sqrt{2 \cdot {5}^{2}} = 5 \sqrt{2}$.

By definition, a radian is the angle subtended by an arc of length equal to the radius. Thus, an arc which subtends an angle of $\theta$ has a length of $\theta r$. In this case, $\theta = \frac{\pi}{8}$, so arc length = $\frac{5 \sqrt{2} \pi}{8}$.

Hint to remember
Note that the circumference subtends and angle of $2 \pi$ at the center, and has a length of $2 \pi r$. By unitary method, an arc which subtends an angle of $\theta$ has a length of $\theta r$.