Points (2 ,9 ) and (1 ,3 ) are (3 pi)/4 radians apart on a circle. What is the shortest arc length between the points?

1 Answer
P dilip_k
Mar 6, 2016

6.24 unit

Explanation:

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It is evident from the above figure that shortest arcAB having end point A(2,9) and B (1,3) will subtend pi/4 rad angle at the center O of the circle. AB chord is obtained by joining A,B. A perpendicular OC is also drawn on it at C from center O.
Now the triangle OAB is isosceles having OA=OB=r (radius of circle)
Oc bisects /_AOB and /_AOC becomes pi/8.
AgainAC= BC=1/2AB=1/2*sqrt((2-1)^2+(9-3)^2)=1/2sqrt37

:.AB=sqrt37

Now AB=AC+BC=rsin/_AOC+rsin/_BOC=2rsin(pi/8)
r=1/2AB*(1/sin(pi/8))=1/2sqrt37csc(pi/8)

Now,
Shortest Arc length of AB = Radius*/_AOB=r*/_AOB=r*(pi/4)=1/2sqrt37csc(pi/8)*(pi/4)=6.24unit

**More easily by properties of triangle **
r/sin(3pi/8)=(AB)/sin(pi/4)
r=(AB)/sin(pi/4)*(sin(3pi/8))=sqrt2AB*sin(3pi/8)
Now
Shortest Arc length of AB = Radius*/_AOB=r*/_AOB=r*(pi/4)=sqrt2AB*sin(3pi/8)*pi/4=6.24 unit

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