1. **Stating the problem:** We are given the center of a circle at coordinates $(2a, a - 7)$ and a point on the circle $(11, -9)$. The diameter of the circle is $10\sqrt{2}$ units. We need to find the value(s) of $a$. 2. **Formula and important rules:** The radius $r$ of the circle is half the diameter, so: $$r = \frac{10\sqrt{2}}{2} = 5\sqrt{2}$$ The distance between the center and any point on the circle equals the radius. Using the distance formula between the center $(2a, a-7)$ and the point $(11, -9)$: $$\sqrt{(11 - 2a)^2 + (-9 - (a - 7))^2} = r$$ 3. **Set up the equation:** $$\sqrt{(11 - 2a)^2 + (-9 - a + 7)^2} = 5\sqrt{2}$$ Simplify inside the square root: $$\sqrt{(11 - 2a)^2 + (-2 - a)^2} = 5\sqrt{2}$$ 4. **Square both sides to eliminate the square root:** $$(11 - 2a)^2 + (-2 - a)^2 = (5\sqrt{2})^2$$ Calculate the right side: $$(5\sqrt{2})^2 = 25 \times 2 = 50$$ 5. **Expand the left side:** $$(11 - 2a)^2 = 121 - 44a + 4a^2$$ $$(-2 - a)^2 = 4 + 4a + a^2$$ Sum: $$121 - 44a + 4a^2 + 4 + 4a + a^2 = 50$$ 6. **Combine like terms:** $$121 + 4 + (4a^2 + a^2) + (-44a + 4a) = 50$$ $$125 + 5a^2 - 40a = 50$$ 7. **Bring all terms to one side:** $$5a^2 - 40a + 125 - 50 = 0$$ $$5a^2 - 40a + 75 = 0$$ Divide entire equation by 5: $$a^2 - 8a + 15 = 0$$ 8. **Solve the quadratic equation:** Using the quadratic formula $a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a=1$, $b=-8$, $c=15$: $$a = \frac{8 \pm \sqrt{(-8)^2 - 4 \times 1 \times 15}}{2} = \frac{8 \pm \sqrt{64 - 60}}{2} = \frac{8 \pm \sqrt{4}}{2}$$ $$a = \frac{8 \pm 2}{2}$$ So, $$a = \frac{8 + 2}{2} = 5$$ or $$a = \frac{8 - 2}{2} = 3$$ **Final answer:** The values of $a$ are $3$ and $5$.