Let A(2,3),B(3,-1),and C(-3,2) be three point | vedclass

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(D) Let $O(h, k)$ be the centre of the circle passing through $A(2,3)$,$B(3,-1)$,and $C(-3,2)$.Since $O$ is the centre,$OA = OB = OC$ (radii of the ci

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(D) Let $O(h, k)$ be the centre of the circle passing through $A(2,3)$,$B(3,-1)$,and $C(-3,2)$.Since $O$ is the centre,$OA = OB = OC$ (radii of the circle).$OA^2 = OC^2 \Rightarrow (h-2)^2 + (k-3)^2 = (h+3)^2 + (k-2)^2$$h^2 - 4h + 4 + k^2 - 6k + 9 = h^2 + 6h + 9 + k^2 - 4k + 4$$-4h - 6k = 6h - 4k$$10h + 2k = 0 \Rightarrow k = -5h \quad ... (i)$$OA^2 = OB^2 \Rightarrow (h-2)^2 + (k-3)^2 = (h-3)^2 + (k+1)^2$$h^2 - 4h + 4 + k^2 - 6k + 9 = h^2 - 6h + 9 + k^2 + 2k + 1$$-4h - 6k + 13 = -6h + 2k + 10$$2h - 8k = -3 \quad ... (ii)$Substitute $(i)$ into $(ii)$:$2h - 8(-5h) = -3$$2h + 40h = -3$ $\Rightarrow 42h = -3$ $\Rightarrow h = -\frac{1}{14}$$k = -5h = -5(-\frac{1}{14}) = \frac{5}{14}$Now,calculate $2k - 4h$:$2(\frac{5}{14}) - 4(-\frac{1}{14}) = \frac{10}{14} + \frac{4}{14} = \frac{14}{14} = 1$

Let $A(2,3)$,$B(3,-1)$,and $C(-3,2)$ be three points. If the centre of the circle passing through $A$,$B$,and $C$ is $O(h, k)$,then find the value of $2k - 4h$.

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