Let $\mathrm{A}(2,3), \mathrm{B}(3,-1)$ and $\mathrm{C}(-3,2)$ be three points. If the centre of the circle passing through $A, B$ and $C$ is $(h, k)$, then $2 k-4 h=$

• A

  $0$
• B

  $2$
• C

  $-1$
• Correct answer
  D

  $1$

Solution

Let $(\mathrm{h}, \mathrm{k})$ be the centre of the circle Now $A O=C O \Rightarrow A O^2=C O^2$ $\begin{aligned} & \Rightarrow(2-h)^2+(3-k)^2 \\ & =(3+h)^2+(2-k)^2 \\ & \Rightarrow k=-5 h...(i) \end{aligned}$
$\begin{aligned} & \text { and } \mathrm{AO}=\mathrm{BO} \Rightarrow A O^2=B O^2 \\ & \Rightarrow(2-h)^2+(3-k)^2=(3-h)^2+(1+k)^2 \\ & \Rightarrow 2 h-8 k=-3...(ii) \end{aligned}$ after solving (i) \& (ii), we get $\begin{aligned} & h=\frac{-1}{14}, k=\frac{5}{14} \\ & \text { so, } 2 k-4 h=\frac{10}{14}+\frac{4}{14}=1 \end{aligned}$

Hence, the correct answer is option: 'D'.