Find the circumcentre of the triangle whose vertices are given below.
$i\left)\right(-2,3\left)\right(2,-1)$ and $(4,0)$

Using distance formula, we can find circum center of a triangle

It is nothing but is the point of intersection of all the three perpendicular bisector of sides of A

Let co ordinates of circum center be $(x,y)$ distance formula is used

$D_1$ using point $(-2,3)$

$D_1=\sqrt{{(x+2)}^2+{(y-3)}^2}$

$=\sqrt{x^2+4x+4+y^2-6y+9}$

$=\sqrt{x^2+y^2+4x-6y+13}$

$=D_2$ using point $(2,1)$

$D_2=\sqrt{{(x-2)}^2+{(y+1)}^2}$

$=\sqrt{x^2+4x+4+y^2+2y+1}$

$=\sqrt{x^2+y^2+4x+2y+5}$

$D_3$ using point $(4,0)$

$D_3=\sqrt{{(x-4)}^2+{(y-0)}^2}$

$=\sqrt{x^2-8x+16+y^2}$

$=\sqrt{x^2+y^2-8x+16}$

As $(x,y)$ is equidistant from all three vertices

$D_1=D_2=D_3$

$\therefore D_1=D_2$

$\therefore \sqrt{x^2+y^2+4x-6y+13}=\sqrt{x^2+y^2+4x+2y+5}$

$\therefore x^2+y^2+4x-6y+13=x^2+y^2+4x+2y+5$

$\therefore -6y+13=2y+5$

$\therefore 8=8y$

$\therefore y=1$

$D_1=D_3$

$\therefore \sqrt{x^2+y^2+4x-6y+13}=\sqrt{x^2+y^2-8x+16}$

$\therefore x^2+y^2+4x-6y+13=x^2+y^2-8x+16$

$\therefore 12x=3+6y$

But $y=1$

$\therefore x=\frac{9}{12}=\frac{3}{4}$

$\therefore$Circum center is $(\frac{3}{4},1)$