Question 79·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

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In the $xy$ -plane, a circle passes through the points $(1,1)$ , $(5,1)$ , and $(3,5)$ .

Which choice is the area of the circle?

A chord joins two points on a circle. The center lies on its perpendicular bisector, the line through the chord’s midpoint at a right angle. For a horizontal chord, that gives the center’s $x$ -coordinate; a Desmos regression can find its $y$ -coordinate by matching distances to the given points. Don’t assume the chord is a diameter unless it passes through the center.

Hints

Hint 1
The two points with $y=1$ form a horizontal chord. Its middle tells you the center’s $x$ -coordinate, but not its $y$ -coordinate. What is that $x$ -coordinate?
Hint 2
Every radius, a segment from the center to the circle, has the same length. Give the center an unknown $y$ -coordinate, then set its squared distances to a lower point and the upper point equal.
Hint 3
Once you know the center, use a given point to find the squared radius. The circle’s area is $A=\pi r^2$ , so you can use that squared value directly.

Step-by-step

Use symmetry and equal distances
Step 1
Place the center on the chord’s middle line
The points $(1,1)$ and $(5,1)$ form a horizontal chord. Its midpoint has $x=\frac{1+5}{2}=3$ , so the center must lie on the vertical line $x=3$ . Call the center $(3,k)$ . Don’t set $k=1$ : the chord’s midpoint would be the center only if this chord were a diameter.
Step 2
Match the squared distances
A radius runs from the center to any point on the circle, so $(1,1)$ and $(3,5)$ must be equally far from $(3,k)$ . The first point is $2$ units away horizontally; the second has no horizontal gap. Square and add the horizontal and vertical changes: $4+(1-k)^2=(5-k)^2$ .Equal distances from the center, not the midpoint of any two circle points, determine its location.
Step 3
Find the center’s height in Desmos
Type $4+(1-k)^2\sim(5-k)^2$ . The $\sim$ tells Desmos to fit the unknown $k$ so the distances match. Under PARAMETERS, Desmos shows $k=2.5$ , so the center is $(3,2.5)$ .
Step 4
Find the squared radius
The point $(3,5)$ is directly above the center, so its vertical distance gives the radius. Type $(5-k)^2$ on the next line. Desmos prints $6.25$ , which is $r^2$ ; you don’t need to take its square root for the area formula.
Step 5
Give the area in exact form
Type $\pi(5-k)^2$ to calculate $A=\pi r^2$ . Desmos shows about $19.635$ . For the exact form used in the choices, the previous line gave $r^2=6.25=\frac{25}{4}$ , so the circle’s area is $\frac{25\pi}{4}$ . Choice B.

Question 79·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

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