# Intersections of a Circle with the Four Quadrants

The Ariel University Center in Samaria (Israel) runs an online mathematical olympiad for college students. During the 2009-2010 season they offered several beautiful problems, one of which might appear harder for college students than for younger mathematicians:

Point P = P(a, b) is located in the first quadrant. Consider a circle centered at the point P with radius greater than a² + b², and denote the area of the part of this circle located in the j-th quadrant (j = 1, 2, 3, 4) by Sj . Find S1 - S2 + S3 - S4.

The applet below illustrates the problem. (The circle is draggable. Its radius could be changed with the scrollbar at the bottom of the applet.)

### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

 What if applet does not run?

Solution

Point P = P(a, b) is located in the first quadrant. Consider a circle centered at the point P with radius greater than a² + b², and denote the area of the part of this circle located in the j-th quadrant (j = 1, 2, 3, 4) by Sj . Find S1 - S2 + S3 - S4.

### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

 What if applet does not run?

### Solution

Draw two additional lines x = 2a and y = 2b. Together with the axes, the lines split the circle into four areas: four curvilinear rectangles, or cut-off circular segments (equal two-by-two), four curvilinear triangles (all of which are equal - by symmetry), and a rectangle in the middle centered at point P. In the (algebraic) sum S1 - S2 + S3 - S4 all areas cancel out, except for the middle rectangle. The rectangle has dimensions 2a×2b and the area 4ab.