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?

1 Answer
Jan 27, 2016

( 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 (x_1, y_1) and (x_2, y_2) on the Euclidean plane is given by d = sqrt( (x_1 - x_2)^2 + (y_1 - y_2)^2 ) .
Thus, radius r = sqrt( (7 - 2)^2 + (6 - 1)^2 ) = sqrt(5^2 + 5^2) = sqrt(2 * 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 = pi / 8 , so arc length = ( 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 .