# Points (5 ,4 ) and (2 ,2 ) are (5 pi)/4  radians apart on a circle. What is the shortest arc length between the points?

Feb 11, 2018

the shortest arc happens to be $s = 4.957$

#### Explanation:

Given:
$\left(5 , 4\right)$ is a point on the circle
$\left(2 , 2\right)$is a point on the circle

Angle subtended at the center is $\frac{5 \pi}{4}$ which is reflex lying on the major arc
The minor arc happens to be $2 \pi - \frac{5 \pi}{4} = \frac{3 \pi}{4}$

The triangle formed by the two points and the center happens to be an isoceles triangle with the shortest angle formed at the center being $\alpha = \frac{3 \pi}{4}$.

Remaining angle is
$\left(\pi - \frac{3 \pi}{4}\right) = \frac{\pi}{4}$

Angle at each of the vertex at the base is $\theta = \frac{1}{2} \frac{\pi}{4} = \frac{\pi}{8}$
Mid point happens to be
$\left(\frac{5 + 2}{2} , \frac{4 + 2}{2}\right) = \left(3.5 , 3\right)$

We calculate the radius of circle by pythagoras theorem in the right angled triangle formed by one of the point, say $\left(5 , 4\right)$, the mid-point $\left(3.5 , 3\right)$, and the center of the circle.

Distance between one of the points, say $\left(5 , 4\right)$ and the mid point $\left(3.5 , 3\right)$ is given by
${a}^{2} = {\left(3.5 - 5\right)}^{2} + {\left(3 - 4\right)}^{2} = {\left(- 1.5\right)}^{2} + {\left(- 1\right)}^{2}$
$= 2.25 + 1 = 3.25$
${a}^{2} = 3.25$

Also, the line joining $\left(5 , 4\right) \mathmr{and} \left(3.5 , 3\right)$ happens to be the adjacent side and the radius being hypotenuse for the angle at the point $\left(5 , 4\right)$

Consideing the ratio $\cos \theta$,
${\cos}^{2} \theta = {a}^{2} / {r}^{2}$
$\theta = \frac{\pi}{8}$
${a}^{2} = 3.25$

${\cos}^{2} \left(\frac{\pi}{8}\right) = \frac{3.25}{r} ^ 2$
${\cos}^{2} \frac{\pi}{8} = 0.146$
$0.854 = \frac{3.25}{r} ^ 2$
${r}^{2} = \frac{3.25}{0.854} = 3.808$
$r = \sqrt{3.808}$
$r = 1.951$

Hence, the shortest arc happens to be
$s = r \alpha = 1.951 \frac{3 \pi}{4}$
$s = 1.533$
the shortest arc happens to be $s = 4.957$

Feb 11, 2018

Shortest Arc length $s = 4.5946$

#### Explanation: vec(AD) = sqrt(5-2)^2 + (4-2)^2) = 3.6056

$r = \frac{A D}{2 \sin \left(\frac{\theta}{2}\right)} = \frac{3.6056}{2 \sin \left(\frac{5 \pi}{8}\right)} = 1.95$

Since the center angle $\theta$ is more than $\pi$, to get the shortest arc length we must subtract from $\left(2 \pi\right)$

Shortest length of the arc

$s = r \cdot \theta = 1.95 \cdot \left(2 \pi - \left(\frac{5 \pi}{4}\right)\right) = 1.95 \cdot \left(\frac{3 \pi}{4}\right) = 4.5946$

Feb 11, 2018

Shorter arc length between the points is $4.59$ unit.

#### Explanation:

Angle subtended at the center by the arc is ${\theta}_{1} = \frac{5 \pi}{4}$

Angle subtended at the center by the minor arc is

$\theta = 2 \pi - \frac{5 \pi}{4} = \frac{3 \pi}{4} \therefore \frac{\theta}{2} = \frac{3 \pi}{8}$

Distance between two points $\left(5 , 4\right) \mathmr{and} \left(2 , 2\right)$ is

D= sqrt ((x_1-x_2)^2+(y_1-y_2)^2) =sqrt ((5-2)^2+(4-2)^2 or

$D = \sqrt{13} \approx 3.61$ unit. Chord length $l \approx 3.61$ unit.

Formula for the length of a chord is ${L}_{c} = 2 r \sin \left(\frac{\theta}{2}\right)$

where $r$ is the radius of the circle and $\theta$ is the angle

subtended at the center by the chord.

:. 3.61=2*r*sin((3pi)/8) or r = 3.61/(2*sin((3pi)/8) or

$r = \frac{3.61}{1.85} \approx 1.95 \therefore$ Radius of the circle is $1.95$ unit.

Arc length is ${L}_{a} = \cancel{2 \pi} \cdot r \cdot \frac{\theta}{\cancel{2 \pi}} = r \cdot \theta$ or

${L}_{a} = 1.95 \cdot \frac{3 \pi}{4} \approx 4.59$ unit .

Shorter arc length between the points is $4.59$ unit.[Ans]