What value of x would make rq tangent to circle p at point q?

If Q is tangent to circle P, that means that it is perpendicular to the radius at the point of tangency.  This means that ΔPRQ would be a right triangle.  We can use the Pythagorean theorem to solve this.
9² + 12² = (x+9)²
We use x+9 because the unknown segment x would combine with the radius, 9, to give that side of the triangle (the hypotenuse).
81 + 144 = (x+9)(x+9) ---- remember that squared means multiplied by itself
225 = (x+9)(x+9)
Multiply the right hand side:
225 = x² + 9x + 9x + 81
225 = x² + 18x + 81
Subtract 225 from both sides:
225 - 225 = x² + 18x + 81 - 225
0 = x² + 18x - 144
Use the Quadratic Formula to solve:

This gives us

Since -18 - 30 = -48, and a negative number doesn't make sense in this situation, we have 
x = (-18+30)/2 = 12/2 = 6.

The value of x cannot be negative, hence the value of x would make rq tangent to Circle P at point Q is 6

From the given circle geometry, since the line PQ is perpendicular to QR, hence the triangle PQR is a right-angle triangle.

Given the following from the geometry;

Adjacent side = QR = 12Opposite side = PQ = 9Hypotenuse side = PR = 9 + x

Using the pythagoras theorem;

Factorize the result to have;

Since the value of x cannot be negative, hence the value of x would make rq tangent to Circle P at point Q is 6

Learn more on circle geometry here: link

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