Question

what value of x would make \\(\overleftrightarrow{rq}\\) tangent to circle p at point q?
x = \boxed{}
(there is a diagram showing a circle with center p, a tangent line rq touching the circle at q, a secant line from r passing through another point on the circle, with lengths: the segment from r to q is 12, the radius pq is 9, and the segment from r to the other intersection point with the circle is x.)

Explanation:

Step1: Recall tangent - radius property

A tangent to a circle is perpendicular to the radius at the point of tangency. So, triangle \(PRQ\) is a right triangle with right angle at \(Q\), where \(PQ = 9\) (radius), \(RQ=12\), and \(PR=x + 9\) (since the segment from \(R\) to the circle is \(x\) and the radius is \(9\)).

Step2: Apply Pythagorean theorem

In right triangle \(PRQ\), by the Pythagorean theorem, \(PQ^{2}+RQ^{2}=PR^{2}\). Substituting the known values: \(9^{2}+12^{2}=(x + 9)^{2}\).

First, calculate \(9^{2}=81\) and \(12^{2}=144\). Then \(81 + 144=(x + 9)^{2}\), so \(225=(x + 9)^{2}\).

Take the square root of both sides: \(\sqrt{225}=\vert x + 9\vert\). Since \(x\) represents a length, we consider the positive root. So \(15=x + 9\).

Step3: Solve for \(x\)

Subtract \(9\) from both sides of the equation \(15=x + 9\): \(x=15 - 9=6\).

Answer:

\(6\)