In a circle, chords and intersect at point inside the circle. The chords are perpendicular at . The length of is 12, the length of segment is 5, and the length of is 16. Which choice is the radius of the circle?

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Super-Hard SAT Math Question 79 | Free SAT Prep | Aniko

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

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In a circle, chords $\overline{AB}$ and $\overline{CD}$ intersect at point $E$ inside the circle. The chords are perpendicular at $E$ . The length of $\overline{AB}$ is 12, the length of segment $\overline{AE}$ is 5, and the length of $\overline{CD}$ is 16.

Which choice is the radius of the circle?

Start with circle-specific theorems: intersecting chords gives you an equation linking the chord segments, and the chord length gives you a second equation (sum). After you know both pieces of the second chord, avoid messy geometry by placing the intersection at the origin with the perpendicular chords on the axes. Then the center is just the intersection of the perpendicular bisectors (through the chord midpoints), and the radius follows from a simple distance calculation.

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Find the missing piece of chord $\overline{AB}$

You’re given $AB$ and $AE$ . Use subtraction to get $EB$ .

Use the intersecting chords relationship

For two chords intersecting inside a circle, the product of the two pieces of one chord equals the product of the two pieces of the other chord: $AE\cdot EB=CE\cdot ED$ .

Use perpendicular bisectors to locate the center

The center of a circle is on the perpendicular bisector of every chord. If you place the intersection point at the origin with one chord horizontal and the other vertical, the bisectors become simple lines.

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Model the intersecting chords to find $CE$

Let $x$ represent $CE$ . Enter the equation

y = x(16 - x)

and also enter

y = 35

Find the intersection points; the $x$ -values are the possible values of $CE$ .

Compute the midpoint offset for chord $\overline{CD}$

Create a value for the midpoint’s $y$ -coordinate using

m = x - 8

(using either intersection value of $x$ ; one gives $m>0$ , the other gives $m<0$ , but the radius will be the same).

Compute the radius from the center to point $A$

Since $A=(-5,0)$ and the center is $(1,m)$ , enter

R = sqrt((1 - (-5))^2 + (m - 0)^2)

Then select the radius value that matches one of the answer choices.

Step-by-step Explanation

Use the given chord to find the missing segment length

Since $AB=12$ and $AE=5$ , the remaining part is

$EB=12-5=7$ .

Use the intersecting chords theorem to split $\overline{CD}$

For intersecting chords, $AE\cdot EB=CE\cdot ED$ .

$5\cdot 7=CE\cdot ED \Rightarrow CE\cdot ED=35$ .

Also, $CD=16$ , so

$CE+ED=16$ .

Let $CE=x$ . Then $ED=16-x$ , and

\begin{aligned} CE\cdot ED&=35 \\ x(16-x)&=35 \\ x^2-16x+35&=0 \end{aligned}

\begin{aligned} x&=\frac{16\pm\sqrt{16^2-4(35)}}{2} \\ &=\frac{16\pm\sqrt{256-140}}{2} \\ &=\frac{16\pm\sqrt{116}}{2} \\ &=8\pm\sqrt{29} \end{aligned}

So the two segment lengths are $8+\sqrt{29}$ and $8-\sqrt{29}$ (in some order).

Locate the center using perpendicular bisectors

Place $E$ at the origin and make $\overline{AB}$ horizontal.

Then $A=(-5,0)$ and $B=(7,0)$ .

The midpoint of $\overline{AB}$ is

\left(\frac{-5+7}{2},0\right)=(1,0),

so the perpendicular bisector of $\overline{AB}$ is the vertical line $x=1$ .

Make $\overline{CD}$ vertical. With $CE$ and $ED$ on opposite sides of $E$ , the midpoint of $\overline{CD}$ has $y$ -coordinate

\frac{CE-ED}{2}=\frac{(8+\sqrt{29})-(8-\sqrt{29})}{2}=\sqrt{29}.

So the perpendicular bisector of $\overline{CD}$ is the horizontal line $y=\sqrt{29}$ .

Therefore, the center of the circle is at the intersection: $(1,\sqrt{29})$ (or reflected to $(1,-\sqrt{29})$ , which gives the same radius).

Compute the radius as a distance to any point on the circle

Use the distance from the center $(1,\sqrt{29})$ to point $A(-5,0)$ :

\begin{aligned} R^2&=(1-(-5))^2+(\sqrt{29}-0)^2 \\ &=6^2+29 \\ &=65 \end{aligned}

So the radius is $\sqrt{65}$ .

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