What is the common area between the circles `y^2-4y+x^2-2x-11=0` and `x^2 + y^2 = 16`  

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txmedteach | High School Teacher | (Level 3) Associate Educator

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To answer this question, we'll need to first simplify the first equation into the standard equation for a circle (`(x-h)^2 + (y-k)^2 = r^2`). We start off with the following:

`y^2+4y+x^2-2x-11 = 0`

Start by adding 11 to both sides so we can complete the square for the x and y expressions:

`y^2 + 4y+x^2-2x = 11`

Now, to "complete the square," which means to make a perfect-square trinomial (that can then be factored into the standard form for a circle), we just divide the coefficient of the "`x`" and "`y`" terms by 2, square that value, then add it to the equation. We see this here:

`y^2 + 4y + 4 + x^2 - 2x + 1 = 11 + 4 + 1`

Notice that we had to add the same amounts to both sides! Simplifying into the standard form, we get:

`(x-1)^2 + (y+2)^2 = 4^2`

Now, let's look at how these two circles intersect by graphing below:

Now, the best way to do this is to throw in some trigonometry. What we'll do is carve out the sectors involved (whose area will be `A = theta/(2pi)pir^2`  where theta is the angle at the center of the circle defining the sector), but ignore the triangle created by the points of intersection and the center of each circle (which will have a height . What we'll basically have is something that looks like an ice cream cone, but we don't care about the cone part, we just want the ice cream on top. The area of that portion will now be given by the area of the full sector less the area of the triangle. Notice, the height of the triangle will be `rcos(theta/2)` and the base will be `2rsin(theta/2)`

`A = theta/(2pi)pir^2 - 1/2(rcos(theta/2))(2rsin(theta/2)) = theta/(2pi)pir^2 - r^2/2sintheta`

The formula on the right is found by using the sine double angle formula. Now, we need to add these areas originating from both circles in the problem. This means, we need to find the points of intersection and angle `theta` associated with them and the center.

To find points of intersection, we simply solve for `x` and `y`:

`x^2 + y^2 = (x-1)^2 + (y+2)^2`

`x^2 + y^2 = x^2-2x+1+y^2+4y+4` 

`0 = -2x+4y+5`

`y = 1/2x - 5/4`

Now, we know that this line defines our points of intersection on the circles. We substitute to solve for `x`:

`x^2 + (1/2x-5/4)^2 = 16`


`x^2 + 1/4x^2 - 5/4x + 25/16 = 16` 

`x^2 - x + 5/4 = 64/5`

`x^2 - x - 231/20 = 0`

Using the quadratic equation:

`x = 3.94, -2.94`

Using our linear equation (`y = 1/2x - 5/4`) we find the following points of intersection:

`(3.94, 0.72), (-2.94, -2.72)`

Now, we need to find the angle.


To find these angles, we use the following formula to give us the angle from the `x`-axis:

`theta_x = tan^-1(y/x)`

`theta_1 = tan^-1(0.72/3.94) = 0.181`

`theta_2 = tan^(-1)((-2.72)/-2.94) = 0.747`

Looking at the circle, it's clear that `theta_2` is too small. Our angle needs to be between `pi` and `3pi/2`, but we found an angle less than `pi/2`. We actually have a simple solution to this: add `pi`!

`theta_2 = 0.747 + pi = 3.889`

If we subtract these angles, notice that we get the external angle. Thus, we need to subtract their difference from `2pi` to get the `theta` we'll use in our area equation.

`theta = 2pi - (3.889 - 0.181) = 2.575`

Finally, we can use the area equation we derived! Notice that because of symmetry, the angles for both circles are going to be the same (we can just double the area from one circle), giving us the following result:

`A = 2(theta/(2pi)pir^2 - r^2/2sintheta)`

`A = 2(2.575/(2pi)*pi*16 - 16/2*sin(2.575)) = 32.6`

Your final area is 32.6.


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