SAT Question c0b53183: Circles answer and explanation with Desmos solution

It's given that the circle has center $O$ and points $A$ and $U$ lie on the circle. It follows that segments $OA$ and $OU$ are radii of the circle and therefore congruent. It's also given that line segments $AE$ and $UE$ are tangent to the circle at points $A$ and $U$, respectively. Since line segments $AE$ and $UE$ are tangent to the circle from a common external point, point $E$, line segments $AE$ and $UE$ are congruent and form right angles with radii $OA$ and $OU$, respectively. Therefore, quadrilateral $OAEU$ consists of two congruent right triangles: triangle $OAE$ and triangle $OUE$. It's given that the distance between point $O$ and point $E$ is $510$ centimeters and the distance between point $O$ and point $U$ is $78$ centimeters. It follows that the area of triangle $OUE$ can be found by first using the Pythagorean theorem to calculate the distance between point $U$ and point $E$. For right triangle $OUE$, where angle $U$ is a right angle, the Pythagorean theorem states that $(OU{)}^2+(UE{)}^2=(OE{)}^2$. Substituting $78$ for $OU$ and $510$ for $OE$ in this equation yields $78^2+(UE{)}^2=510^2$, or $6,084+(UE{)}^2=260,100$. Subtracting $6,084$ from both sides of this equation yields $(UE{)}^2=254,016$. Taking the positive square root of both sides of this equation yields $UE=504$. Thus, triangle $OUE$ is a right triangle with legs of length $78$ centimeters and $504$ centimeters. It follows that the area of triangle $OUE$ is $(\frac{1}{2})(78)(504)$ square centimeters, or $19,656$ square centimeters. Since triangle $OAE$ is congruent to triangle $OUE$, the area of triangle $OAE$ is also $19,656$ square centimeters. Therefore, the area of quadrilateral $OAEU$ is $19,656+19,656$ square centimeters, or $39,312$ square centimeters.