Circles

**classicstrings** · Nov 13th 2006, 06:58 PM

The area of the intersection of two circles of radius 5 cm which have their centres 8 cm apart is?

I get a right angled triangle with two sides length 5cm and a hyp of 8. I work out the angle between the 5 side and 8 side, then times that by 2 to get the full angle.

Then use the formula, segment = .5r^2(theta - sintheta), then times that answer by two to get the final answer (the interection) which is wrong.

**ThePerfectHacker** · Nov 13th 2006, 07:28 PM

I will assume the circles do not pass the centers.

Look a diagram below.

To find the intersecting area find the half area, i.e. the shaded region.

Which can be found by finding the full area of the sector and subtracting the full area of the triangle.

The area of the triangle is simple $\frac{1}{2}\cdot 4\cdot 6=12$

The area of the sector is more complicated. It is required to find the central angle. To do this we can use the formula $\frac{1}{2}\cdot 5^2\cdot \sin t=12$ that is another formula for triangle area (half the product of the sides and sine of included angle). Thus, $t\approx 74^o$ .

Thus the sector area is,
$\frac{74^o}{360^o}\pi (5)^2\approx 16$

The shades reagion is,
$16-12\approx 4$
Thus, the full intersection area is $\approx 8$

**Soroban** · Nov 13th 2006, 07:46 PM

Hello, classicstrings!

Careful . . . 5-5-8 is not a right triangle.
. . But your game plan is correct.

The area of the intersection of two circles of radius 5 cm
which have their centres 8 cm apart is?

Code:

                  C
                  *
               *  :  *
           5*     :3    *5
         * α      :        *
      *  *  *  *  *  *  *  *  *
     A      4           4      B

I get a triangle with sides 5, 5, and 8.

We also get a couple of 3-4-5- right triangles.

I work out the angle between the 5 side and 8 side.