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(C) Let the centre of the circle be $(h, k)$. Since the circle passes through $(3, -2)$ and $(-2, 0)$,the distances from the centre to these points ar

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$x^2 + y^2 + 3x + 12y + 2 = 0$

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(C) Let the centre of the circle be $(h, k)$. Since the circle passes through $(3, -2)$ and $(-2, 0)$,the distances from the centre to these points are equal (radius squared):$(h - 3)^2 + (k + 2)^2 = (h + 2)^2 + (k - 0)^2$$h^2 - 6h + 9 + k^2 + 4k + 4 = h^2 + 4h + 4 + k^2$$-6h + 4k + 13 = 4h + 4$$10h - 4k = 9$  --- $(1)$The centre $(h, k)$ lies on the line $2x - y = 3$,so:$2h - k = 3$  --- $(2)$Multiplying $(2)$ by $4$,we get $8h - 4k = 12$. Subtracting this from $(1)$:$(10h - 4k) - (8h - 4k) = 9 - 12$$2h = -3 \Rightarrow h = -\frac{3}{2}$Substituting $h$ into $(2)$:$2(-\frac{3}{2}) - k = 3 \Rightarrow -3 - k = 3 \Rightarrow k = -6$The centre is $(-\frac{3}{2}, -6)$. The radius squared $r^2$ is:$r^2 = (-\frac{3}{2} + 2)^2 + (-6 - 0)^2 = (\frac{1}{2})^2 + 36 = \frac{1}{4} + 36 = \frac{145}{4}$The equation is $(x + \frac{3}{2})^2 + (y + 6)^2 = \frac{145}{4}$$x^2 + 3x + \frac{9}{4} + y^2 + 12y + 36 = \frac{145}{4}$$x^2 + y^2 + 3x + 12y + \frac{153}{4} = \frac{145}{4}$$x^2 + y^2 + 3x + 12y + 2 = 0$.

The equation of the circle which passes through the points $(3, -2)$ and $(-2, 0)$ and whose centre lies on the line $2x - y = 3$ is:

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