Question

applying the converse of the radius-tangent theorem
what value of x would make \overleftrightarrow{rq} tangent to circle p at point q?
x = \square

Explanation:

Step1: Recall Radius-Tangent Theorem

If a line is tangent to a circle, the radius to the point of tangency is perpendicular to the tangent line. So, \( \triangle PQR \) is a right triangle with \( \angle PQR = 90^\circ \), \( PQ = 9 \) (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 \( PQR \), \( PQ^2 + RQ^2 = PR^2 \). Substitute the known values: \( 9^2 + 12^2 = (x + 9)^2 \).

Calculate \( 9^2 = 81 \), \( 12^2 = 144 \), so \( 81 + 144 = (x + 9)^2 \).

\( 225 = (x + 9)^2 \). Take square roots: \( \sqrt{225} = x + 9 \), so \( 15 = x + 9 \).

Step3: Solve for \( x \)

Subtract 9 from both sides: \( x = 15 - 9 = 6 \).

Answer:

\( x = 6 \)