Question 50·SAT Desmos Regression Practice·Geometry and Trigonometry

From $A$ to $B$, the coordinate changes are $(7-1,4-1)=(6,3)$. From $A$ to $C$, they’re $(4-1,-5-1)=(3,-6)$. Swapping the changes and reversing one sign turns $(6,3)$ into $(3,-6)$, a quarter-turn. So the two segments meet at a right angle at $A$.

All three beacons are on the circle, so the right angle at $A$ is an inscribed angle: its vertex lies on the circle. The side opposite an inscribed right angle is a diameter. So $BC$ is the diameter, not either side that meets at $A$.

From $B(7,4)$ to $C(4,-5)$, the coordinate changes are $-3$ and $-9$. The Pythagorean theorem adds their squares to get the squared diameter. Since the radius is half the diameter, divide its square by $4$. Type $\frac{(4-7)^2+(-5-4)^2}{4}$ in Desmos. It displays $22.5$; tap the fraction button to see $r^2=\frac{45}{2}$.

Area is the space inside the circular boundary, so use $A=\pi r^2$. Substitute the squared radius directly:

Multiply by $\pi$:

The circle’s area is $\frac{45\pi}{2}$ square units. Choice B.

If you miss the right angle, three points can still determine the circle. Enter their coordinates in a Desmos table, then type $x_1^2+y_1^2\sim ux_1+vy_1+w$. The $\sim$ asks Desmos to fit the constants. This is the expanded circle equation: for center $(h,k)$, $u=2h$, $v=2k$, and $w=r^2-h^2-k^2$. Under PARAMETERS, Desmos shows $u=11$, $v=-1$, and $w=-8$.

Because $u$ and $v$ are twice the center’s coordinates, add $h=u/2$ and $k=v/2$. Desmos displays $h=5.5$ and $k=-0.5$, so the center is $(5.5,-0.5)$.

The fitted constant is $w=r^2-h^2-k^2$, so add the center’s squares back. Type $q=w+h^2+k^2$, where $q$ names $r^2$. Desmos displays $q=22.5=\frac{45}{2}$.

Type $\pi q$ for the area formula $A=\pi r^2$. Desmos displays about $70.686$; since $q=\frac{45}{2}$, the exact area is $\frac{45\pi}{2}$ square units. Choice B.