The centre of a circle is $(2a, a-7)$. Find the values of $a$ if the circle passes through the point $(11, -9)$ and has a diameter of $10\sqrt{2}$ units.

(A-D) Given that the centre of the circle is $C(2a, a-7)$ and it passes through the point $P(11, -9)$.
The distance between the centre $C$ and the point $P$ on the circle is equal to the radius $r$.
Using the distance formula,$r = \sqrt{(11 - 2a)^2 + (-9 - (a - 7))^2} = \sqrt{(11 - 2a)^2 + (-2 - a)^2}$.
The diameter of the circle is $10\sqrt{2}$ units,so the radius $r = \frac{10\sqrt{2}}{2} = 5\sqrt{2}$ units.
Equating the two expressions for the radius:
$5\sqrt{2} = \sqrt{(11 - 2a)^2 + (-2 - a)^2}$.
Squaring both sides:
$(5\sqrt{2})^2 = (11 - 2a)^2 + (-2 - a)^2$
$50 = (121 - 44a + 4a^2) + (4 + 4a + a^2)$
$50 = 5a^2 - 40a + 125$
$5a^2 - 40a + 75 = 0$.
Dividing by $5$:
$a^2 - 8a + 15 = 0$
$a^2 - 5a - 3a + 15 = 0$
$a(a - 5) - 3(a - 5) = 0$
$(a - 5)(a - 3) = 0$.
Therefore,the values of $a$ are $a = 5$ or $a = 3$.