Problem Solving

Set T consists of all points (x, y) such that x^2 + y^2 = 1 . If point (a, b) is selected from set T at random, what is the probability that b > (a + 1) ?

a) 1/4
b) 1/3
c) 1/2
d) 3/5
e) 2/3

[spoiler]
OA is a, can somebody help me visualize this problem?[/spoiler]

gmatclub problem

Set T consists of all points (x, y) such that x^2 + y^2 = 1 . If point (a, b) is selected from set T at random, what is the probability that b > (a + 1) ?

x^2 + y^2 = 1 is a circle of radius 1 and center as origin.

Now, we need to find, how many points exist on this circle that have b>a+1
Of the four quadrants that the points exist, the points with b>a+1 cannot exist on the positive side of x-axis.
because, point (a+1) will be out of the circle.

Now, we are left with 2 quadrants where x is negative.
Again, if both x and y are negative, then also b>a+1 will fail.
So we are left with only 1 quadrant.

Any point on the circle in this quadrant will have b>a+1. for example (-root(3)/2 , 1/2) (-1/root(2),1/root(2)),etc..

Hence the probability is 1/4.

Hope this helps!!

x^2 + y^2 = 1 is a circle of radius 1 with the origin as its center.

You want to know the probability that the y coordinate is more than 1 greater than the x coordinate (i.e. b-a > 1).

This comes down to examining the four quadrants:

Quadrant I - this involves all points along circle between (0,1) and (1,0). For these points, b-a is 1 and -1, respectively. The value of b-a will range between -1 and 1, and thus never be bigger than 1. No points in Q1 fit.

Quadrant II - this involves all points along circle between (0,1) and (-1,0). For these points, b-a is 1 and 1, respectively. So for all points in the range, b-a is either always > 1 or always <1. For (-root(2), root(2)), the difference between b and a is greater than 1, so all points in Q2 fit the range.

Quadrant III - this involves all points along circle between (-1,0) and (0,-1). For these points, b-a is 1 and -1, respectively. Therefore b-a ranges from -1 to 1 and thus cannot be > 1. No points from Q3 fit.

Quadrant IV - this involves all points along circle between (0,-1) and (1,0). For these points, b-a is -1 and -1, respectively. For (root(2), -root(2)), the difference between b and a is less than -1, so b-a < -1 for all points in Q4. No points in Q4 fit.

Only those points in Q2 fit. This comprises 1/4 of the circumference of the circle, and thus the probability that b > a+1 is 1/4.

Make sense?