Super-Hard SAT Math Question 141 | Free SAT Prep | Aniko

00:00

Question 141·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

0 / 200

In the $xy$ -plane, circle $C$ lies entirely in the first quadrant and is tangent to both the $x$ -axis and the $y$ -axis. The center of the circle is inside the square with vertices $(0,0)$ , $(20,0)$ , $(20,20)$ , and $(0,20)$ . The line $x+y=14$ intersects circle $C$ at two points $A$ and $B$ .

If the length of chord $\overline{AB}$ is $8$ , which choice is the radius of circle $C$ ? Thi‍ѕ queѕtіon іѕ frоm ⁠Aniko

When a circle is tangent to both coordinate axes in the first quadrant, immediately rewrite its center as $(r,r)$ and its radius as $r$ to reduce unknowns. Then treat the line that creates the chord as a chord line: compute the perpendicular distance from the center to that line, and use the right-triangle relationship $r^2=d^2+(\text{half-chord})^2$ . Solve the resulting quadratic and use the extra condition (here, the center inside the $20\times 20$ square) to select the correct solution. Pr‌opertу of Ani⁠k‍о.аі

Hints

Click me!

Use the tangency information

If a circle in the first quadrant is tangent to both axes, write its center and radius using a single variable $r$ . ЅAТ рreр by Аniko.аi

Connect the chord to a distance from the center

For a chord cut by a line, the key distance is the perpendicular distance from the circle’s center to that line (here, $x+y=14$ ).

Use a right triangle with half the chord

A radius to the midpoint of a chord forms a right triangle whose legs are the center-to-line distance and half the chord length.

Desmos Guide

Set up the two expressions using $x$ as the radius variable

Let $x$ represent the radius. Enter

y1 = x^2

and

y2 = ((14 - 2x)^2)/2 + 16

Find intersection $x$ -values

Graph both. The valid radius values are the $x$ -coordinates where $y1$ and $y2$ intersect (you can click the intersection points to read the coordinates).

Apply the square condition

Choose the intersection with $0<x<20$ so the center $(x,x)$ lies inside the $20\times 20$ square. Aniк‌ο.аі - ⁠ЅAТ ‌Рrеp

Step-by-step Explanation

Use tangency to place the center

If circle $C$ is tangent to both the $x$ -axis and the $y$ -axis and lies in the first quadrant, then its radius is $r>0$ and its center is

(r,r).

Find the distance from the center to the chord’s line

The chord $\overline{AB}$ lies on the line $x+y=14$ , i.e., $x+y-14=0$ .

The perpendicular distance from $(r,r)$ to this line is

d=\frac{|r+r-14|}{\sqrt{1^2+1^2}}=\frac{|2r-14|}{\sqrt2}.

(We can keep the absolute value because the next step uses $d^2$ .) Fro‌m ‌aniкo.аi

Relate chord length, radius, and distance to the chord

A radius drawn to the midpoint of chord $\overline{AB}$ is perpendicular to the chord, forming a right triangle.

Half the chord length is $\frac{8}{2}=4$ . With hypotenuse $r$ and one leg $d$ , we have

r^2=d^2+4^2=d^2+16.

Substitute and solve, then choose the radius that fits the square

Substitute $d^2=\left(\frac{|2r-14|}{\sqrt2}\right)^2=\frac{(2r-14)^2}{2}$ into $r^2=d^2+16$ :

\begin{aligned} r^2 &= \frac{(2r-14)^2}{2}+16. \end{aligned}

Multiply by $2$ and simplify:

\begin{aligned} 2r^2 &= (2r-14)^2+32 \\ 2r^2 &= (4r^2-56r+196)+32 \\ 0 &= r^2-28r+114. \end{aligned}

r=\frac{28\pm\sqrt{28^2-4\cdot 114}}{2}=\frac{28\pm\sqrt{328}}{2}=14\pm\sqrt{82}.

The value $14+\sqrt{82}$ is greater than $20$ , so its center $(r,r)$ would not be inside the given $20\times 20$ square. Therefore the radius is $14-\sqrt{82}$ .

Strategy

Hints

Desmos Guide

Step-by-step Explanation

Study Hints

Strategy

Hints

Desmos Guide

Step-by-step Explanation