In the -plane, a circle has center at and passes through point . A second point on the circle is such that measures (the angle is measured at the center of the circle). What is the length of ?

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SAT Circles Question 38 | Free SAT Prep | Aniko

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Question 38·Hard·Circles

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In the $xy$ -plane, a circle has center $O$ at $(4,-2)$ and passes through point $P(10,1)$ . A second point $Q$ on the circle is such that $\angle POQ$ measures $150^\circ$ (the angle is measured at the center of the circle).

What is the length of $\overline{PQ}$ ?

For circle questions involving a chord and a central angle, first find the radius using the distance formula (if needed), then view the radii and the chord as forming an isosceles triangle. Drop a perpendicular from the center to the chord to create two right triangles, and use $\frac{\text{chord}}{2} = r\sin(\tfrac{\text{central angle}}{2})$ so that $\text{chord} = 2r\sin(\tfrac{\text{central angle}}{2})$ . Finally, evaluate any special-angle trig values (like $75^\circ = 45^\circ + 30^\circ$ ) using exact radicals to match the answer choices quickly.

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Start with the radius

Use the distance formula to find the distance from the center $O(4,-2)$ to the point $P(10,1)$ . That distance is the radius of the circle.

Think about triangle $POQ$

$OP$ and $OQ$ are both radii of the circle, and the angle between them at $O$ is $150^\circ$ . How does the side $PQ$ relate to this triangle?

Relate the chord to half the angle

If you drop a perpendicular from the center $O$ to the chord $PQ$ , it splits $\angle POQ$ into two equal angles and splits $PQ$ in half. Which trigonometric function connects half of $PQ$ , the radius, and half of the central angle?

Use exact trig values

Once you have an expression involving $\sin 75^\circ$ , write $75^\circ$ as $45^\circ + 30^\circ$ and use exact values for $\sin 45^\circ, \cos 45^\circ, \sin 30^\circ$ , and $\cos 30^\circ$ .

Desmos Guide

Compute the radius

Enter the expression sqrt((10-4)^2 + (1-(-2))^2) in Desmos to confirm the radius $r$ of the circle. Note the exact or decimal value that Desmos gives.

Set up the chord-length expression

In a new line, type 2 * sqrt((10-4)^2 + (1-(-2))^2) * sin(150°/2) to represent $PQ = 2r\sin(150^\circ/2) = 2r\sin 75^\circ$ . Make sure to include the degree symbol so Desmos uses degrees.

Compare with the answer choices

Look at the value Desmos gives for this expression. Then, for each answer choice, type its expression into Desmos (for example, 3*sqrt(5), 3*sqrt(2), etc.) and compare their decimal values to the value you found for $PQ$ . The choice that matches is the correct answer.

Step-by-step Explanation

Find the radius of the circle

The radius is the distance from the center $O(4,-2)$ to point $P(10,1)$ .

Using the distance formula:

OP = \sqrt{(10-4)^2 + (1-(-2))^2} = \sqrt{6^2 + 3^2} = \sqrt{36 + 9} = \sqrt{45} = 3\sqrt{5}.

So the radius of the circle is $r = 3\sqrt{5}$ .

Relate the chord $PQ$ to triangle $POQ$

Points $P$ and $Q$ lie on the circle, and $O$ is the center, so $OP = OQ = r = 3\sqrt{5}$ . Thus, $\triangle POQ$ is isosceles with sides $OP$ and $OQ$ equal and central angle $\angle POQ = 150^\circ$ .

The segment $PQ$ is the chord opposite the $150^\circ$ angle at $O$ .

Express $PQ$ in terms of $r$ and the central angle

Drop a perpendicular from $O$ to the chord $PQ$ ; this perpendicular bisects both the chord and the angle $\angle POQ$ .

So each right triangle has:

hypotenuse $r$ ,
acute angle $75^\circ$ (half of $150^\circ$ ),
and half of the chord $PQ$ as the side opposite $75^\circ$ .

Thus,

\frac{PQ}{2} = r\sin 75^\circ \quad\Rightarrow\quad PQ = 2r\sin 75^\circ.

Substitute $r = 3\sqrt{5}$ to get

PQ = 2(3\sqrt{5})\sin 75^\circ.

Evaluate $\sin 75^\circ$ and simplify

Use the angle-sum identity with $75^\circ = 45^\circ + 30^\circ$ :

\sin 75^\circ = \sin(45^\circ + 30^\circ) = \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ.

Now plug in exact values:

\sin 45^\circ = \frac{\sqrt{2}}{2},\quad \cos 30^\circ = \frac{\sqrt{3}}{2},\quad \cos 45^\circ = \frac{\sqrt{2}}{2},\quad \sin 30^\circ = \frac{1}{2},

\sin 75^\circ = \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2}\cdot\frac{1}{2} = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4}.

Substitute this into $PQ = 2(3\sqrt{5})\sin 75^\circ$ :

PQ = 6\sqrt{5}\cdot\frac{\sqrt{6} + \sqrt{2}}{4} = \frac{3\sqrt{5}}{2}(\sqrt{6} + \sqrt{2}).

Now simplify the radicals:

\sqrt{5}\sqrt{6} = \sqrt{30},\quad \sqrt{5}\sqrt{2} = \sqrt{10},

PQ = \frac{3}{2}\bigl(\sqrt{30} + \sqrt{10}\bigr),

which matches answer choice D.

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