The centre of a circle is (2a, a - 7). Find the value (values) of a, if the circle passes through the point (11, -9) and has diameter $10{\sqrt2}$ units.

Radius of circle = $\dfrac{\text{Diameter}}{2}$

= $\dfrac{10 \sqrt2}{2}$

= $5 \sqrt2$

⇒ Distance between the given points = $\sqrt{(x$2 - x1)^2 + (y2 - y1)^2}

Radius of the circle = The distance between the center (2a, a - 7) and the point (11, -9), which lies on the circle

$5 \sqrt2 = \sqrt{(2a - 11)^2 + ((a - 7) - (-9))^2}\\[1em] ⇒ (5 \sqrt2)^2 = (2a - 11)^2 + (a - 7 + 9)^2\\[1em] ⇒ (5 \sqrt2)^2 = (2a - 11)^2 + (a + 2)^2\\[1em] ⇒ 50 = 4a^2 + 121 - 44a + a^2 + 4 + 4a\\[1em] ⇒ 50 = 5a^2 + 125 - 40a\\[1em] ⇒ 5a^2 + 125 - 40a - 50 = 0\\[1em] ⇒ 5a^2 - 40a + 75 = 0\\[1em] ⇒ a^2 - 8a + 15 = 0\\[1em] ⇒ a^2 - (3a + 5a) + 15 = 0\\[1em] ⇒ a^2 - 3a - 5a + 15 = 0\\[1em] ⇒ a(a - 3) - 5(a - 3) = 0\\[1em] ⇒ (a - 3)(a - 5) = 0\\[1em] ⇒ a = 3 \text{ or } 5$

Hence, the value of a = 3 or a = 5.