Question

what value of $x$ would make $overleftrightarrow{rq}$ tangent to circle p at point q?
$x = \square$
length of $rq$ is 12, length of $pq$ is 9, length of the segment marked as $x$ is the difference between $rp$ and the radius of the circle p

Explanation:

Step1: Recognize right triangle formation

If $\overleftrightarrow{RQ}$ is tangent to circle $P$ at $Q$, then $\angle PQR = 90^\circ$, so $\triangle PQR$ is a right triangle with hypotenuse $PR = x + 9$, leg $PQ = 9$, and leg $RQ = 12$.

Step2: Apply Pythagorean theorem

$$(x + 9)^2 = 9^2 + 12^2$$
Calculate the right-hand side:
$$9^2 + 12^2 = 81 + 144 = 225$$
So:
$$(x + 9)^2 = 225$$

Step3: Solve for $x$

Take square roots of both sides:
$$x + 9 = 15$$
(We discard the negative root since length is positive)
Subtract 9 from both sides:
$$x = 15 - 9$$

Answer:

$6$