What is the value of x given that segment RQ is tangent to circle P at point Q?

Question

Answer 1

We use the Pythagorean combined with a little algebra to solve this.

First, let us set up our formula:

1. c = squareroot ( a²+b² ), where a and be are the legs of the right triangle
2. Let's start by defining a, b and c, then we can work on solving for x:
3. a = x
4. b = 9
5. c = (x+4)

Now, let's set up our formula:

1. (x+4) = squareroot ( x² + 9² )
2. To solve for x we need to eliminate the square root, we do this by squaring both sides of our equation:
3. (x+4)² = x² + 9²
4. Expand the left side using FOIL:
5. x² + 8x + 16 = x² + 9²
6. Combine like terms:
7. 8x = 9² - 16 = 65
8. Solve for x
9. x = 65/8

Answer 2

If the segment is tangent to the circle, then the triangle must be a right triangle.

So, by the Pythagorean Theorem, x² + 9² = (x+4)²

x² + 81 = x² + 8x + 16

8x + 16 = 81

8x = 65

x = 65/8