Circle and square

Since $BF$ is tangent to the circle, by Tangent-Secant Theorem , $\begin{aligned} \;\;\;\;\;BF^2&=BE\cdot BC \\ x^2&=BE\cdot (10+x) \\ \implies BE&=\frac{x^2}{10+x} \end{aligned}$ In $\triangle COG$ , $\begin{aligned} OC^2&=CG^2+GC^2 \\ 10^2&=x^2+x^2 \\ \implies x&=\sqrt{50} \end{aligned}$ Area of triangle $ABE$ is, $\begin{aligned} \Delta&=\frac{1}{2}\cdot AB\cdot BE \\ &=\frac{1}{2}\cdot (10+x)\cdot \frac{x^2}{10+x} \\ &=\frac{x^2}{2} \\ &=\boxed{25} \end{aligned}$

Let the center of the circle be $O$ and its radius $r$ , $AB$ is tangent to the circle at $F$ , and the side length of square $ABCD$ be $a$ . Then $a = \left(1+\dfrac 1{\sqrt 2}\right) r$ .

Let $EB = d$ . Then by tangent-secant theorem , $EB \cdot BC = FB^2 \implies da = (a-r)^2$ , and the area of $\triangle ABE$ .

$\begin{aligned} A_\triangle & = \frac {da}2 = \frac {(a-r)^2}2 = \frac {\left(\left(1+\frac 1{\sqrt 2}-1\right)r\right)^2}2 = \frac {r^2}4 = \frac {10^2}4 = \boxed {25} \end{aligned}$

for $r =10$ .

Refer to the picture from the solution from Sathvik Acharya. Let L = AB = CB (they are equal since they are both sides of the same square). Construct radius OE (which, like all radii of the circle, has length 10). Then, by the hypotenuse-leg condition, we see triangles OCG and OEG are congruent. Therefore, GE has length x so that EB has length:

EB = L - 2*x. (equation 1)

But, by the Pythagorean Theorem, 2*x^2 = 100, so that x = sqrt(50)

However, we also note that 10 + x = DC = L, so that L = 10 + sqrt(50).

Thus, by substitution into equation 1, EB = 10 -sqrt(50)

We then note that:

Area (AEB) = (1/2) L (EB) = (1/2) (10 + sqrt(50)) (10 - sqrt(50)) = 50/2 = 25

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