Question

applying the converse of the radius - tangent theorem. what value of x would make rq tangent to circle p at point q? x =

Explanation:

Step1: Recall the radius - tangent property

If $\overrightarrow{RQ}$ is tangent to circle $P$ at point $Q$, then $\angle PQR = 90^{\circ}$ and triangle $PQR$ is a right - triangle.

Step2: Apply the Pythagorean theorem

In right - triangle $PQR$, by the Pythagorean theorem $PR^{2}=PQ^{2}+RQ^{2}$. Here, $PQ = 9$, $RQ = 12$, and $PR=x + 9$. So, $(x + 9)^{2}=9^{2}+12^{2}$.

Step3: Calculate the right - hand side

$9^{2}+12^{2}=81 + 144=225$. So, $(x + 9)^{2}=225$.

Step4: Solve for $x$

Take the square root of both sides: $x + 9=\pm15$. Since $x$ represents a length and must be non - negative, we consider the positive root. $x+9 = 15$, then $x=15 - 9=6$.

Answer:

$6$