# A Difficult Area Problem involving a Circle and a Square

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.

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 $$AB$$.

• Thanks for the answer! Still wondering, though, if there is a quicker way to find the length $AB$ i.e. without the need to get into coordinate geometry. Apr 2, 2021 at 13:26
• Yes there is but it will still require to solve equations. I will edit. Apr 2, 2021 at 13:29
• Okay perfect, thanks 😊 Apr 2, 2021 at 13:39