1. **Problem statement:** Given a circle with two tangent lines meeting outside the circle, the radius to the tangent points is $x$, and the base between tangent points is 6. Also, inside the circle, a right triangle has one leg 3 and hypotenuse $x$. Find $x$. 2. **Key properties:** - Tangent segments from a common external point are equal in length. - The radius to a tangent point is perpendicular to the tangent line. - In the right triangle inside the circle, use the Pythagorean theorem. 3. **From the bottom-left diagram:** The right triangle has leg 3 and hypotenuse $x$, so by Pythagoras: $$x^2 = 3^2 + b^2$$ where $b$ is the other leg. 4. **From the top-left diagram:** The base between tangent points is 6, and the two radii to tangent points are $x$ each. The triangle formed is isosceles with sides $x, x, 6$. 5. **Using the Pythagorean theorem on the isosceles triangle:** Drop a perpendicular from the center $P$ to the base 6, splitting it into two segments of length 3 each. The height $h$ satisfies: $$h^2 + 3^2 = x^2$$ 6. **Relate $h$ to the right triangle inside the circle:** From step 3, $x^2 = 9 + b^2$. From step 5, $h^2 + 9 = x^2$. Equate: $$h^2 + 9 = 9 + b^2$$ $$h^2 = b^2$$ So $h = b$. 7. **Conclusion:** The height $h$ equals the other leg $b$ of the right triangle. 8. **Find $x$ using the right triangle:** Since $x^2 = 9 + b^2$ and $h = b$, and $h^2 + 9 = x^2$, substitute $h = b$: $$x^2 = 9 + h^2$$ But also from the isosceles triangle: $$x^2 = h^2 + 9$$ Both expressions are consistent. 9. **Final step:** Given the base is 6 and one leg of the right triangle is 3, the hypotenuse $x$ is: $$x = \sqrt{3^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$$ **Answer:** $$x = 3\sqrt{2}$$