[[1]] is the circumcenter and circumradius of the triangle whose vertices are (1, 1), (2, -1), and (3, 2).

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Now we are given that the vertices of triangle are (1, 1), (2, -1) and (3, 2)
Let us say A = (1, 1), B = (2, -1) and C = (3, 2).

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Now let O = (x, y) be the circumcenter of the triangle.
Now we can say that since O is circumcenter, OA = OB = OC = r.
Now Let us first calculate distance OA
We know that distance between two points (x1,y1)(x1,y1) and (x2,y2)(x2,y2) is given by

= \sqrt{{(x_{2} - x_{1})}^{2} - {(y_{2} - y_{1})}^{2}}

r = \sqrt{{(x - 1)}^{2} - {(y - 1)}^{2}}

r^{2} = {(x - 1)}^{2} - {(y - 1)}^{2}

Now using

{(a - b)}^{2} = a^{2} + b^{2} - 2 ab

we get

r^{2} = x^{2} + 1 - 2 x + y^{2} + 1 - 2 y

...........(1)
Now Let us calculate the distance OB.
We know that distance between two points (x1,y1) and (x2,y2) is given by

= \sqrt{{(x_{2} - x_{1})}^{2} - {(y_{2} - y_{1})}^{2}}

r = \sqrt{{(x - 2)}^{2} - {(y - (- 1))}^{2}}

r^{2} = {(x - 2)}^{2} - {(y + 1)}^{2}

Now using

{(a - b)}^{2} = a^{2} + b^{2} - 2 ab

and

{(a + b)}^{2} = a^{2} + b^{2} + 2 ab

we get

r^{2} = x^{2} + 4 - 4 x + y^{2} + 1 + 2 y . . . . . . . . . . . (2)

And finally we calculate distance OC.
We know that distance between two points (x1,y1) and (x2,y2) is given by

= \sqrt{{(x_{2} - x_{1})}^{2} - {(y_{2} - y_{1})}^{2}}

r = \sqrt{{(x - 3)}^{2} - {(y - 2)}^{2}}

r^{2} = {(x - 3)}^{2} - {(y - 2)}^{2}

Now using

{(a - b)}^{2} = a^{2} + b^{2} - 2 ab

we get

r^{2} = x^{2} + 9 - 6 x + y^{2} + 4 - 4 y . .

.........(3)
Now equating equation (1) and equation (2) we get

x^{2} + 1 - 2 x + y^{2} + 1 - 2 y = x^{2} + 4 - 4 x + y^{2} + 1 + 2 y

⇒1−2x+1−2y=4−4x+2y+1
⇒2−2x−2y=5−4x+2y
⇒4x−2x−2y−2y=5−2
∴ 2x−4y=3.......................................(4)
And again equating equation (1) and equation (3) we get.

x^{2} + 1 - 2 x + y^{2} + 1 - 2 y = x^{2} + 9 - 6 x + y^{2} + 4 - 4 y

⇒1−2x+1−2y=9−6x+4−4y
⇒6x−2x+4y−2y=9+4−1−1
∴ 4x+2y=11......................................(5)
Now multiplying equation (5) by 2 and adding the equation to equation (1) we get
8x+4y+2x−4y=22+3
⇒10x=25
∴ x= 2.5
Now substituting x = 2.5 in equation (4) we get.
2(2.5)−4y= 3
⇒5−3= 4y
⇒4y= 2

\Rightarrow y = \frac{1}{2} = 0.5

∴ y= 0.5
Hence we have x = 2.5 and y = 0.5.
Hence we get the coordinates of circumcenter is (2.5, 0.5)
Now we just need to find the circumradius.
Circumradius is nothing but the distance between circumcenter and any vertex. Let us take this vertex as A = (1, 1) for solving
Then we know that distance between two points (x1,y1) and (x2,y2) is given by