Question 22·25 Super-Hard SAT Geometry and Trigonometry Questions·Geometry and Trigonometry

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In the $xy$ -plane, circle $T$ has its center on the line $y=x$ and contains the points $(0,6)$ and $(8,0)$ . Circle $U$ is obtained by reflecting circle $T$ across the $x$ -axis and then doubling its radius. The equation of circle $U$ can be written as

(x-a)^2+(y-b)^2=p,

where $a$ , $b$ , and $p$ are constants. What is the value of $p$ ?

Your answer

A circle’s center is equidistant, or equally far, from every point on its edge. If the center lies on a line and two points lie on the circle, use the line to describe the center, then equate its squared distances to the points. A Desmos regression can find the unknown coordinate. Don’t assume the two points are diameter endpoints, and remember that doubling a radius quadruples its square.

Hints

Hint 1
The line $y=x$ contains points whose two coordinates match. Give the center one unknown coordinate and use it for both coordinates.
Hint 2
Every point on a circle is one radius from its center. The two given points must have equal squared distances from the center. What equation do you get by squaring and adding each point’s horizontal and vertical gaps?
Hint 3
The number on the right side of a circle equation is the radius squared, not the radius. A reflection keeps the radius the same; what happens to its square when the radius doubles?

Step-by-step

Find the center with equal distances
Step 1
Describe the center using the line
The center lies on $y=x$ , so its two coordinates must match. Call their common value $t$ ; the center of circle $T$ is $(t,t)$ .
Step 2
Set the two squared distances equal
Both $(0,6)$ and $(8,0)$ are on circle $T$ , so each is one radius from $(t,t)$ . Squaring and adding the horizontal and vertical gaps gives their squared distances, which must match:
$(t-0)^2+(t-6)^2=(t-8)^2+(t-0)^2.$
Don’t use the points’ midpoint as the center. The problem never says they’re ends of a diameter.
Step 3
Find the center coordinate in Desmos
Type the equation with $t$ replaced by $x_1$ and $=$ replaced by $\sim$ . Desmos uses $x_1$ as the unknown to fit, and the regression symbol $\sim$ tells it to find a value that makes the sides equal. Under PARAMETERS, Desmos shows $x_1=7$ , so circle $T$ has center $(7,7)$ .
Step 4
Find circle T’s squared radius
Let $r_T$ be circle $T$ ’s radius. The distance from its center $(7,7)$ to $(0,6)$ is already one radius, so don’t halve it. Type $(x_1-0)^2+(x_1-6)^2$ . Desmos prints $50$ , which means $r_T^2=50$ .
Step 5
Account for the doubled radius
Reflecting circle $T$ across the $x$ -axis doesn’t change its radius. Circle $U$ then doubles that radius, so the right side of its equation is:
$p=(2r_T)^2.$
Square the $2$ :
$p=4r_T^2.$
Doubling a radius makes its square four times as large. Type $4\cdot50$ ; Desmos prints $200$ . Since $p$ is the squared radius of circle $U$ , Grid in 200.

Question 22·25 Super-Hard SAT Geometry and Trigonometry Questions·Geometry and Trigonometry

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