Real 2013 SAT math problem

One student in my math club shared this real math problem. Thank you.

This is a simple problem for most students. However, there are several concepts I think are important for students.



Let's take a look at the properties of two tangent circles. There are two positions of tangent circles. See figure below.

Two circles with centers at O1, O2 and with radium r1, r2. Then, the left one witch is externally tangent circles, we should have the length of O1O2 = r1+r2; the right one which is internally tangent circles, we should have O102 = |r1-r2|.

Solution
Base on the above property, AB = 2r, and we know BC = r. Also, because AC is perpendicular to BC, by the Pythagorean Theorem, AC^2 = AB^2 - BC^2 = 4r^2 - r^2 = 3r^2. Therefore, AC is SQR(3)*r. So, (C) is the correct answer.

Furthermore, I want to review the Tangent to a circle properties:
A tangent to a circle is perpendicular to the radius at the point of tangency.

• This is a very useful property when the radius that connects to the point of tangency is part of a right angle, because the trigonometry and the Pythagorean Theorem apply to right triangles.

The tangent segments to a circle from an external point are equal.

The angle between a tangent and a chord is equal to the inscribed angle on the opposite side of the chord.