The centre of a circle is (2x - 1, 3x + 1). Find x if the circle passes through (-3, -1) and the length of its diameter is 20 units.

The diameter of the circle is given as 20 units, so the radius is 10 units.

Distance between the centre A (2x - 1, 3x + 1) and point B (-3, -1) = Radius of circle AB = 10

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

⇒ AB² = 100

Distance between B(-3, -1) and A(2x - 1, 3x + 1):

$⇒ ((2x - 1) - (-3))^2 + ((3x + 1) - (-1))^2 = 100\\[1em] ⇒ (2x - 1 + 3)^2 + (3x + 1 + 1)^2 = 100\\[1em] ⇒ (2x + 2)^2 + (3x + 2)^2 = 100\\[1em] ⇒ 4x^2 + 4 + 8x + 9x^2 + 4 + 12x = 100\\[1em] ⇒ 13x^2 + 8 + 20x = 100\\[1em] ⇒ 13x^2 + 8 + 20x - 100 = 0\\[1em] ⇒ 13x^2 + 20x - 92 = 0\\[1em] ⇒ x = \dfrac{-20 + \sqrt{400 + 4784}}{26} \text { or } \dfrac{-20 - \sqrt{400 + 4784}}{26}\\[1em] ⇒ x = \dfrac{-20 + \sqrt{5184}}{26} \text { or } \dfrac{-20 - \sqrt{5184}}{26}\\[1em] ⇒ x = \dfrac{-20 + 72}{26} \text { or } \dfrac{-20 - 72}{26}\\[1em] ⇒ x = \dfrac{52}{26} \text { or } \dfrac{-92}{26}\\[1em] ⇒ x = 2 \text { or } \dfrac{-46}{13}\\[1em]$

Hence, the values of x are 2 and $\dfrac{-46}{13}$.