Let $C_1(O_1, r)$ and $C_2(O_2, r)$ be two congruent secant circles. Calculate in function of the $r$, the area of the quadrilateral $AO_1BO_2$.

Question

Let $C_1(O_1, r)$ and $C_2(O_2, r)$ be two congruent secant circles, so that $C_1 \cap C_2$ = {A, B} and $O_1 \in C_2, O_2 \in C_1$. Calculate in function of the radius, the area of the quadrilateral $AO_1BO_2$....

MY IDEAS.

My drawing: (the wrong drawing)

THE RIGHT DRAWING

Okey, so the first thing I have thought of is the theorem that calculates the area of a quadrilateral using $r$.

A()= p*r where p=semiperimeter

Then I realised that $AO_1BO_2$ is a rhombus because $AO_1=O_1B=O_2B=O_2A=$Radius of the circles (because the 2 circles are congruent)

I don't know what else to do forward. All ideas are welcome! Thank you!!!

Answer 1

Let $R$ be the radius of the inscribed circle.

Let $C$ be the tangent point of the inscribed circle with $AO_2$.

Let $O'$ be the center of the inscribed circle.

Since $\angle{O'CA}=90^\circ$, we have $$\text{Area}(O'AO_2)=\frac 12\times O'C\times AO_2=\frac{rR}{2}$$

Similarly, we have $$\text{Area}(O'AO_1)=\text{Area}(O'O_1B)=\text{Area}(O'BO_2)=\frac{rR}{2}$$

Therefore, the area of the quadrilateral $AO_1BO_2$ is $$4\times\frac{rR}{2}=\color{red}{2rR}$$

---

Added 1 :

I realized that I did not use the condition that $O_1 \in C_2, O_2 \in C_1$ from which we can say that $O_1O_2=r$.

So, we can say that $\triangle{AO_1O_2}$ is an equilateral triangle, so we have $$\text{Area}(AO_1O_2)=\frac 12\times r\times \frac{\sqrt 3}{2}r=\frac{\sqrt 3}{4}r^2$$

Therefore, the area of $AO_1BO_2$ is $$2\times\frac{\sqrt 3}{4}r^2=\color{red}{\frac{\sqrt 3}{2}r^2}$$

---

Added 2 :

why $O_1O_2$ equals ths radius?

The condition $O_1 \in C_2$ means that $O_1$ is on the circle $C_2$.

The condition $O_2 \in C_1$ means that $O_2$ is on the circle $C_1$.

It follows from these that $O_1O_2=r$.

that means my drawing is made wrong?

Your drawing is not correct.

$O_1$ has to be on the circle $C_2$, and $O_2$ has to be on the circle $C_1$.

Answer 2

The function that describes the area of rhombus $AO_1BO_2$ is $$\begin{align}f()&=\frac 12O_1O_2\times AB\\&=\frac 12O_1O_2\times2\sqrt{r^2-\left(\frac{O_1O_2}2\right)^2}\\f(x)&=\frac 12x\sqrt{(2r)^2-x^2}\\x&=O_1O_2, 0\lt x\lt 2r\\f\prime(x)&=0\Rightarrow x=r\sqrt2\end{align}$$

Therefore, the area of rhombus $AO_1BO_2$ is described by the function $$f:(0, 2r)\to (0, r\sqrt 2), f(x)=\frac 12x\sqrt{(2r)^2-x^2}$$ with a maximum of $$f(r\sqrt 2)=r^2\frac{\sqrt{2}}2$$

In terms of the radius on inner circle and the radii of the given congruent circles, the problem is elementary since the rhombus is also a parallelogram and the area becomes $$side\times height=r\times 2r_{\text{inner circle}}$$