A circle has a center at (1 ,2 ) and passes through (4 ,7 ). What is the length of an arc covering pi /4  radians on the circle?

Jul 10, 2016

Length of Arc$= \sqrt{34} \cdot \frac{\pi}{4} \cong 4.58 u n i t$

Explanation:

Let us denote by $S$ the Circle, by $r$ radius, and pt.$C \left(1 , 2\right)$= Centre of $S$, and a pt $P \left(4 , 7\right)$ on $S$

Then, dist. $C P = r ,$ and, using Dist. Formula, we get,

${r}^{2} = C {P}^{2} = {\left(4 - 1\right)}^{2} + {\left(7 - 2\right)}^{2} = 9 + 25 = 34$, giving $r = \sqrt{34}$

The arc subtends angle $\theta = \frac{\pi}{4}$ at the centre, so that,

Length of Arc$= r \cdot \theta = \sqrt{34} \cdot \frac{\pi}{4} \cong 5.84 \cdot \frac{3.14}{4} = 4.58 u n i t$