1. **State the problem:** We have a circle centered at point $Q$ with a vertical diameter line passing through points $A$, $Q$, and $B$. The top and bottom horizontal chords are each 16 units long. 2. The perpendicular distances from the center $Q$ to the top and bottom chords are given as $4x + 3$ and $7x - 6$ respectively. 3. **Goal:** Solve for the radius of the circle. 4. **Formula and rules:** For a chord of length $c$ at a perpendicular distance $d$ from the center of a circle with radius $r$, the relationship is: $$r^2 = d^2 + \left(\frac{c}{2}\right)^2$$ This comes from the right triangle formed by the radius, the perpendicular distance to the chord, and half the chord length. 5. Since the circle has two chords of equal length 16, and their distances from the center are $4x + 3$ and $7x - 6$, both satisfy the formula: $$r^2 = (4x + 3)^2 + 8^2$$ $$r^2 = (7x - 6)^2 + 8^2$$ 6. Set the two expressions for $r^2$ equal to each other: $$(4x + 3)^2 + 64 = (7x - 6)^2 + 64$$ 7. Cancel 64 from both sides: $$(4x + 3)^2 = (7x - 6)^2$$ 8. Expand both sides: $$(4x)^2 + 2 \cdot 4x \cdot 3 + 3^2 = (7x)^2 - 2 \cdot 7x \cdot 6 + 6^2$$ $$16x^2 + 24x + 9 = 49x^2 - 84x + 36$$ 9. Rearrange to one side: $$0 = 49x^2 - 84x + 36 - 16x^2 - 24x - 9$$ $$0 = 33x^2 - 108x + 27$$ 10. Simplify by dividing all terms by 3: $$0 = 11x^2 - 36x + 9$$ 11. Solve quadratic equation $11x^2 - 36x + 9 = 0$ using the quadratic formula: $$x = \frac{36 \pm \sqrt{(-36)^2 - 4 \cdot 11 \cdot 9}}{2 \cdot 11}$$ Calculate discriminant: $$(-36)^2 - 4 \cdot 11 \cdot 9 = 1296 - 396 = 900$$ 12. Calculate roots: $$x = \frac{36 \pm 30}{22}$$ Two solutions: $$x_1 = \frac{36 + 30}{22} = \frac{66}{22} = 3$$ $$x_2 = \frac{36 - 30}{22} = \frac{6}{22} = \frac{3}{11}$$ 13. Check which $x$ gives positive distances: For $x=3$: $4x + 3 = 4(3) + 3 = 15$ (positive) $7x - 6 = 7(3) - 6 = 15$ (positive) For $x=\frac{3}{11}$: $4x + 3 = 4 \cdot \frac{3}{11} + 3 = \frac{12}{11} + 3 = \frac{45}{11} \approx 4.09$ (positive) $7x - 6 = 7 \cdot \frac{3}{11} - 6 = \frac{21}{11} - 6 = \frac{21 - 66}{11} = -\frac{45}{11} \approx -4.09$ (negative, invalid) So $x=3$ is the valid solution. 14. Calculate radius $r$ using $x=3$ and the formula: $$r^2 = (4x + 3)^2 + 8^2 = 15^2 + 64 = 225 + 64 = 289$$ $$r = \sqrt{289} = 17$$ **Final answer:** The radius of the circle is $17$ units.