The circumcentre of the triangle with vertices $(-2,3)$, $(2,-1),(4,0)$ is
The circumcentre of the triangle with vertices $(-2,3)$, $(2,-1),(4,0)$ is
$\left(\frac{3}{2}, \frac{5}{2}\right)$
$\left(\frac{3}{2}, \frac{-5}{2}\right)$
$\left(\frac{-3}{2}, \frac{5}{2}\right)$
$\left(\frac{-3}{2}, \frac{-5}{2}\right)$
Solution
Given the vertices $(-2,3),(2,-1), \&(4,0)$
Let $O(x, y)$ be the circumcentre of the triangle.
Let $\triangle \mathrm{ABC}$ be the triangle.
Then $\mathrm{OA}=\mathrm{OB}=\mathrm{OC}$
$\Rightarrow \mathrm{OA}^2 \mathrm{OB}^2 \Rightarrow(\mathrm{x}+2)^2+(\mathrm{y}-3)^2=(\mathrm{x}-2)^2+(\mathrm{y}+1)^2$
$\begin{aligned}
& \Rightarrow x-y=-1 ... (i)\\
& \text { and } O C^2=O B^2 \\
& \Rightarrow(x-4)^2+(y-0)^2=(x-2)^2+(y+1)^2 \\
& \Rightarrow 4 x+2 y=11 ... (ii)
\end{aligned}$
By solving (i) \& (ii), we get $x=\frac{3}{2}, y=\frac{5}{2}$