1. **Problem statement:** We have a circle with center P and two radii each of length $x$ forming a right angle. A tangent line touches the circle at the endpoint of one radius, creating a right triangle with the radius and a segment of length 6 outside the circle. We need to find the value of $x$. 2. **Key fact:** The radius to the point of tangency is perpendicular to the tangent line, so the triangle formed is right-angled at the tangent point. 3. **Identify the triangle sides:** The hypotenuse is $x$ (distance from center P to the tangent point), one leg is the radius $x$, and the other leg outside the circle is 6. 4. **Apply the Pythagorean theorem:** For the right triangle with legs $x$ and 6, and hypotenuse $x$, we have $$x^2 = x^2 + 6^2$$ 5. **Simplify the equation:** $$x^2 = x^2 + 36$$ 6. **Subtract $x^2$ from both sides:** $$\cancel{x^2} = \cancel{x^2} + 36 \implies 0 = 36$$ This is a contradiction, so the hypotenuse cannot be $x$ if the other leg is 6 and one leg is $x$. 7. **Re-examine the problem:** The triangle formed has legs $x$ (radius) and 6 (tangent segment), and the hypotenuse is the segment from P to the point on the tangent line, which is $x + x = 2x$ (since the hypotenuse is longer than the radius). 8. **Apply Pythagorean theorem correctly:** $$ (2x)^2 = x^2 + 6^2 $$ $$ 4x^2 = x^2 + 36 $$ 9. **Solve for $x$:** $$ 4x^2 - x^2 = 36 $$ $$ 3x^2 = 36 $$ $$ x^2 = \frac{36}{3} = 12 $$ $$ x = \sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3} $$ 10. **Final answer:** $$ x = 2 \sqrt{3} $$