Question 122·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

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In circle $O$ , chord $\overline{AB}$ has length $10$ . The perpendicular distance from the center $O$ to chord $\overline{AB}$ is $5$ . The lines tangent to the circle at $A$ and at $B$ intersect at point $P$ outside the circle.

Which choice gives the length of $\overline{PA}$ ?

A chord is a segment joining two points on a circle. A perpendicular from the center cuts a chord in half, so compare that half with the center-to-chord distance to look for a special right triangle. Then use the right angle between a radius and a tangent to find the angles in the triangle containing the requested length. Don’t mistake the chord for a tangent segment.

Hints

Hint 1
A perpendicular from the center to a chord cuts the chord into equal halves. Call the meeting point $M$ . What is $AM$ when the whole chord is $10$ ?
Hint 2
Compare the two legs, the sides forming the right angle, in triangle $OAM$ . Equal legs mean its two acute angles are equal. What is the angle at $A$ ?
Hint 3
A radius to a point of tangency meets the tangent at a right angle. Find the angle between $PA$ and $AB$ , then use the equal tangents from $P$ to identify triangle $PAB$ .

Step-by-step

Find the hidden 45-45-90 triangle
Step 1
Halve the chord
No Desmos needed. Special-right-triangle ratios give exact lengths here. Let $M$ be where the perpendicular from $O$ meets the chord $AB$ . A perpendicular from a circle’s center bisects a chord, so $M$ is its midpoint: $AM=MB=\frac{10}{2}=5$ .
Step 2
Find the angle at A
The given distance makes $OM=5$ , so triangle $OAM$ has a right angle at $M$ and equal legs, $OM=AM=5$ . Equal legs in a right triangle make both acute angles $45^\circ$ . So $\angle OAM=45^\circ$ .
Step 3
Transfer the angle to the tangent
A tangent touches the circle at one point, and the radius $OA$ is perpendicular to tangent $PA$ there. Since $AM$ lies along $AB$ , the angle between $PA$ and $AB$ is the part of that right angle left after $\angle OAM$ : $\angle PAB=90^\circ-45^\circ=45^\circ$ .
Step 4
Match the base angles
Tangents drawn from the same outside point are equal, so $PA=PB$ . Triangle $PAB$ is isosceles, meaning its angles across from those equal sides match: $\angle PBA=\angle PAB=45^\circ$ .
Step 5
Identify the hypotenuse
The angles in a triangle sum to $180^\circ$ . With $45^\circ$ at both $A$ and $B$ , angle $P$ is $90^\circ$ . So $AB$ is the hypotenuse, the side opposite that right angle, of triangle $PAB$ .
Step 6
Find the tangent length
In a $45^\circ$ - $45^\circ$ - $90^\circ$ triangle, the hypotenuse is either leg times $\sqrt2$ . $AB=10$ is the hypotenuse, so: $PA\sqrt2=10$ . Divide by $\sqrt2$ : $PA=\frac{10}{\sqrt2}$ . Multiply top and bottom by $\sqrt2$ : $PA=\frac{10\sqrt2}{2}$ . Simplify: $PA=5\sqrt2$ . The length of tangent segment $PA$ is $5\sqrt2$ . Choice B.

Question 122·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

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