Question

what value of x would make $overleftrightarrow{rq}$ tangent to circle p at point q?
$x = square$

Explanation:

Step1: Recall tangent - radius property

If $\overleftrightarrow{RQ}$ is tangent to circle $P$ at point $Q$, then $\angle PQR = 90^{\circ}$ (a tangent to a circle is perpendicular to the radius at the point of tangency). So, triangle $PQR$ is a right - triangle.

Step2: Apply Pythagorean theorem

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

Step3: Calculate the right - hand side

First, calculate $9^{2}+12^{2}$. We know that $9^{2}=81$ and $12^{2}=144$, so $9^{2}+12^{2}=81 + 144=225$.

Step4: Expand and solve for $x$

Expand $(x + 9)^{2}=x^{2}+18x + 81$. So, $x^{2}+18x + 81=225$. Rearrange to get a quadratic equation: $x^{2}+18x+81 - 225=0$, which simplifies to $x^{2}+18x - 144=0$.
We can also solve from $(x + 9)^{2}=225$. Take the square root of both sides: $x + 9=\pm15$.
Case 1: $x+9 = 15$, then $x=15 - 9=6$.
Case 2: $x + 9=-15$, then $x=-15 - 9=-24$. But since $x$ represents a length, we discard the negative value.

Answer:

$6$