Problem involving the square root of a trigonometric term

Question

I was trying to find the shaded area in this figure:

And no, it isn't homework. I just chanced upon it on Facebook and had a go at it.

I managed to find it using a very simple method. I now want to verify that my answer is correct with calculus. Specifically, I want to verify my answer with integration on an area bounded between two curves which are the smaller and larger circle.

To do this, I did the following to frame the problem more simply:

Rearranging the larger circle to make it simpler down the line

Then, as you can see in the image, the two polar equations of the circles were constructed. I actually had no idea how to do it for circles that are not centered on the origin or on any part of the x or y axes, so I referred to this link to get the equation for that:

Plotting polar equations of circles not centered at (0, 0)

The next step is to equate the two equations so as to find out where they intersect. This is what I got:

$5\sqrt2\cos(\theta-\frac{\pi}{4}) + \sqrt{5^2-50\sin^2(\theta-\frac{\pi}{4})} = 10$

$5\sqrt2\cos(\theta-\frac{\pi}{4}) + 5\sqrt{1-2\sin^2(\theta-\frac{\pi}{4})} = 10$

$5\sqrt2\cos(\theta-\frac{\pi}{4}) + 5\sqrt{\cos(2(\theta-\frac{\pi}{4}))} = 10$

$\sqrt2\cos(\theta-\frac{\pi}{4}) + \sqrt{\cos(2(\theta-\frac{\pi}{4}))} = 2$

And I am stuck here pretty much. I did a bit of thinking and realized that I am not really sure how to solve this type of equation. Checking on Wolfram Alpha, I could find the value of theta. They are as follows:

https://www.wolframalpha.com/input/?i=10+%3D+sqrt(50)cos(theta+-+pi%2F4)+%2B+sqrt(25+-+50(sin(theta+-+pi%2F4))%5E2)

https://www.wolframalpha.com/input/?i=sqrt(2)cos(theta+-+pi%2F4)+%2B+sqrt(cos(2(theta+-+pi%2F4)))+%3D+2

I did try solving it in Cartesian form then converting it to Polar coordinates later but I didn't manage to solve that either:

$(x-5)^2 + (y-5)^2 = 5^2$

$x^2 + y^2 = 10^2$

With some simple substitution I ended up with:

$x + \sqrt{10^2-x^2} - 5 = 0$

which I also don't know how to solve analytically. I can easily solve any of these numerically but I want to know how one would go about them analytically, preferably without converting them to complex form.

In any case, once the intersections are found, I can perform the necessary integration to obtain the area. That I know I can do.

Thank you!

EDIT:

Okay I must admit I am very tired and this has affected my basic mathematical skills such as rearranging equations /facepalm.

This problem is so simple it is not even worth asking. Apologies for the waste of server space. The trick is to bring the root term to one side and everything else to the other side. Square it and voila, everything becomes easy.

Answer 1

Refer to the circles in rectangular coordinates. Expand the first and substitute for $x^2+y^2,$ to get $$x+y=12.5.$$ This may be substituted into the circle centred at the origin, and the rest should be quite straightforward.

Answer 2

Here are the steps to solve your $\theta$-equation: $$\sqrt2\cos\left(\theta-\frac{\pi}{4}\right) + \sqrt{\cos\left[2(\theta-\frac{\pi}{4})\right]} = 2$$ $$2-\sqrt2\cos\left(\theta-\frac{\pi}{4}\right) = \sqrt{\cos\left[2(\theta-\frac{\pi}{4})\right]} $$ $$\left[2-\sqrt2\cos\left(\theta-\frac{\pi}{4}\right)\right]^2 = \cos\left[2(\theta-\frac{\pi}{4})\right]$$ $$4-4\sqrt2\cos(\theta-\frac{\pi}{4}) + 2\cos^2(\theta-\frac{\pi}{4})=2 \cos^2(\theta-\frac{\pi}{4})-1$$ $$\cos(\theta-\frac{\pi}{4}) =\frac{5}{4\sqrt{2}}$$ $$\theta_1=\pi/4+\cos^{-1}\left(\frac{5}{4\sqrt{2}}\right)=72.9°$$ $$\theta_2=\pi/4-\cos^{-1}\left(\frac{5}{4\sqrt{2}}\right)=17.1°$$

In $xy$-coordinates, you should get the quadratic equation for $x$,

$$8x^2-100x+225=0$$

and the solutions $$x_1=\frac{5}{4}(5-\sqrt{7}), x_2=\frac{5}{4}(5+\sqrt{7})$$