A Difficult Area Problem involving a Circle and a Square

Question

A few days ago, I encountered the following problem:

After a little bit of thinking, I managed to come up with the following solution:

1. Rotate the square $90^\circ$ clockwise and let the new bottom left corner of the square be $(0,0)$.
2. The circle inscribed in the square is hence centered at $(5,5)$ with a radius of $5$. The circle equation thus becomes $(x-5)^{2} + (y-5)^{2} = 25 \Rightarrow y = 5 + \sqrt{25 - (x-5)^{2}}$ in the first quadrant.
3. Similarly for the quarter circle, the equation becomes $y = \sqrt{100-x^2}$.

The graph hence looks like this:

My intention is to find the shaded area in the above graph. To do so, first I find $X$ by equating $5 + \sqrt{25 - (x-5)^{2}} = \sqrt{100-x^2} \Rightarrow x=\frac{25 - 5\sqrt{7}}{4}$.

From this, I calculate the area of the shaded region as follows: $$\text{Area} = (10 \cdot \frac{25 - 5\sqrt{7}}{4} - \int_0^\frac{25 - 5\sqrt{7}}{4} \sqrt{100-x^2} \,\mathrm{d}x) + (10 \cdot (5 - \frac{25 - 5\sqrt{7}}{4}) - \int_\frac{25 - 5\sqrt{7}}{4}^5 5 + \sqrt{25 - (x-5)^{2}} \,\mathrm{d}x) \approx 0.7285$$

Now, the diagram looks like this:

From here, I figured out the shaded area as follows: $$\text{Area} \approx 10^{2} - \frac{\pi(10^{2})}{4} - (\frac{10^{2} - \pi(5^{2})}{4} + 2 \times 0.7285) \approx \boxed{14.6 \: \text{cm}^{2}}$$

While I did figure out the correct solution, I find my approach to be rather lengthy. I was wondering if there is a quicker, simpler and more concise method (that probably does not require Calculus) that one can use and I would highly appreciate any answers pertaining to the same.

Answer 1

Here is an alternate solution,

We assume right bottom vertex to be the origin then, equations of circles are

Circle S: $x^2+y^2 = 100$
Circle T: $(x+5)^2+(y-5)^2 = 25$

Solving both equations, intersection points are $A \left(\frac{5}{4}\left(\sqrt7-5\right), \frac{5}{4}\left(\sqrt7+5\right)\right)$ and $B\left(-\frac{5}{4}\left(\sqrt7+5\right), -\frac{5}{4}\left(\sqrt7-5\right)\right)$

So length of chord $AB$ at intersection is $\frac{5 \sqrt7}{\sqrt2}$.

We note that this chord $AB$ is common chord of both circles.

Angle subtended by chord at the center is given by,

At $P$, $\angle APB = \alpha = 2 \arcsin \left(\frac{\sqrt7}{2\sqrt2}\right)$

At $O$, $\angle AOB = \beta = 2 \arcsin \left(\frac{\sqrt7}{4\sqrt2}\right)$

Shaded area is difference of area of two circular segments of this chord, which are $ATB$ and $ASB$.

$ = \left(circular \ sector \ PATB - \triangle PAB\right) - \left(circular \ sector \ OASB - \triangle OAB\right)$

$ = \left(25 \times \frac{\alpha}{2} - \frac{25 \sqrt7}{8}\right) - \left(100 \times \frac{\beta}{2} - \frac{125 \sqrt7}{8}\right) \approx 14.638 $

EDIT: to find $AB$ without coordinate geometry, we know that $OP = 5 \sqrt2$. If perp from $P$ to $AB$ is $x$ then,

$\left(5\sqrt2 + x\right)^2 + \left(\frac{AB}{2}\right)^2 = 10^2$ and $x^2 + \left(\frac{AB}{2}\right)^2 = 5^2$. Solving them, we get value of $AB$.

Answer 2

The parametric equation for the full circle, assuming the center is at the origin is

$ r_1(t) = ( 5 \cos(t), 5 \sin(t) ) $

And the parametric equation for the quarter circle of radius 10 is

$ r_2(s) = (5 - 10 \sin(s), -5 + 10 \cos(s) ) \hspace{25pt} s \in [0, \dfrac{\pi}{2}]$

Next, we want to find the intersection between $r_1$ and $r_2$. There are obviously two solutions, and these are solutions of the following trigonometric system

$$ 5 \cos(t) = 5 - 10 \sin(s) $$

$$ 5 \sin(t) = -5 + 10 \cos(s) $$

Squaring both equations and adding results in

$ 25 = 50 + 100 - 100 \sin(s) - 100 \cos(s) $

which reduces to

$ \sin(s) + \cos(s) = \dfrac{5}{4} $

Using the angle shift trick, we get

$ \sqrt{2} \cos( s - \dfrac{\pi}{4} ) = \dfrac{5}{4} $

From which

$ s = \dfrac{\pi}{4} \pm \cos^{-1}\left( \dfrac{5}{4\sqrt{2}} \right)$

That is,

$s_1 =\dfrac{\pi}{4} - \cos^{-1}\left( \dfrac{5}{4\sqrt{2}} \right)$

$s_2 =\dfrac{\pi}{4} + \cos^{-1}\left( \dfrac{5}{4\sqrt{2}} \right)$

And from this, and from symmetry, it follow that

$ \cos(s_1) = \dfrac{1}{\sqrt{2}} \cdot \dfrac{5}{4\sqrt{2}} + \dfrac{1}{\sqrt{2}} \cdot \dfrac{\sqrt{7}}{4 \sqrt{2}} = \dfrac{5 + \sqrt{7}}{8} $

And

$ \cos(s_2) = \dfrac{5 - \sqrt{7} }{8} $

$ \sin(s_1) = \cos(s_2) = \dfrac{5 - \sqrt{7}}{8} $

$ \sin(s_2) = \cos(s_1) = \dfrac{5 + \sqrt{7}}{8} $

Using the original equations, we can compute $\cos(t)$ and $\sin(t)$ at these two values of $s$, we have

$ \cos(t_1) = 1 - 2 \sin(s_1) , \sin(t_1) = -1 + 2 \cos(s_1) $

Hence,

$ \cos(t_1) = 1 - \dfrac{5 - \sqrt{7}}{4} = \dfrac{\sqrt{7} - 1 }{4} $

$ \sin(t_1) = -1 + \dfrac{5 + \sqrt{7}}{4} = \dfrac{ \sqrt{7}+1}{4} $

And similarly for $t_2$.

$ \cos(t_2) = \dfrac{-1 - \sqrt{7}}{4}$

$ \sin(t_2) = \dfrac{-\sqrt{7} + 1 }{4} $

Using the $Atan2(x,y)$ function, we can determine the values of $t_1, t_2, s_1, s_2$

$ s_1 = 0.298703208$

$ s_2 = 1.272093118$

$ t_1 = 1.146765287$

$ t_2 = 3.565623693$

So, $s_1, t_1$ correspond to the top intersection point, and $s_2, t_2$ correspond to the left intersection point.

Now we use the shoe-lace formula, which states that

$\text{Area} = \dfrac{1}{2} \displaystyle \bigg| \int_{t_1}^{t_2} x_1(t) y_1'(t) - x_1'(t) y_1(t) \ dt + \int_{s_2}^{s_1} x_2(s) y_2'(s) - x_2'(s) y_2(s) \ d s \bigg| $

Now, from the $x$ and $y$ expressions of $r_1$ and $r_2$

The integrand of the first integral becomes simply $25$, and the second integrand becomes $ 100 - 50(\cos(s) + \sin(s)) $

Therefore, the area computation now becomes,

$\text{Area} = \dfrac{1}{2} \bigg| 25 (t_2 - t_1) + 100 (s_1 - s_2) - 50 ( \sin(s_1) - \sin(s_2) - \cos(s_1) + \cos(s_2) ) \bigg| $

Pluggin the values obtained above, this comes to

$ \text{Area} = \dfrac{1}{2} \bigg| 25 (2.418858406) + 100 (-0.97338991) - 50 (2)(-0.661437828 ) \bigg|$

And finally,

$ \text{Area} \approx 14.638$