Let $\mathrm{A}(2,3), \mathrm{B}(3,-1)$ and $\mathrm{C}(-3,2)$ be three points. If the centre of the circle…
- $0$
- $2$
- $-1$
- $1$
Solution

$\begin{aligned} & \text { and } \mathrm{AO}=\mathrm{BO} \Rightarrow A O^2=B O^2 \\ & \Rightarrow(2-h)^2+(3-k)^2=(3-h)^2+(1+k)^2 \\ & \Rightarrow 2 h-8 k=-3...(ii) \end{aligned}$ after solving (i) \& (ii), we get $\begin{aligned} & h=\frac{-1}{14}, k=\frac{5}{14} \\ & \text { so, } 2 k-4 h=\frac{10}{14}+\frac{4}{14}=1 \end{aligned}$
Asked in: AP EAMCET 2023 (15 May Shift 2)