Question

What value of x would make rq tangent to Circle P at point Q?

Answer 1

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:
$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \\ \\x=\frac{-18 \pm \sqrt{18^2-4(1)(-144)}}{2(1)} \\ \\x=\frac{-18 \pm \sqrt{324--576}}{2} \\ \\x=\frac{-18 \pm \sqrt{324+576}}{2} = \frac{-18 \pm \sqrt{900}}{2}$
This gives us
$x=\frac{-18 \pm 30}{2} = \frac{-18+30}{2} \text{ or } \frac{-18-30}{2}$
Since -18 - 30 = -48, and a negative number doesn't make sense in this situation, we have 
x = (-18+30)/2 = 12/2 = 6.

Answer 2

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 = 12

• Opposite side = PQ = 9

• Hypotenuse side = PR = 9 + x

Using the pythagoras theorem;

$c^2=a^2+b^2\\(9+x)^2=9^2+12^2\\81+18x+x^2=81+144\\x^2+18x+81-225=0\\x^2+18x-144=0$

Factorize the result to have;

$x^2+24x-6x-144=0\\x(x+24)-6(x+24)=0\\(x-6)(x+24) =0\\x=6 \ and \ -24$

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: https://brainly.com/question/22077295