(i) Find the circumcenter and circumradius of the triangle whose vertices are $(1, 1), (2, - 1) a n d (3, 2)$ .

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Solution

Circumcenter of circle whose vertices are
$(1, 1), (2, - 1)$ and $(3, 2)$

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

Let $A = (1, 1) B = (2, - 1) C = (3, 2)$

Let $S (x, y)$ be circumcenter
According to condition
$S A = S B$
$x^{2} + y^{2} - 2 x - 2 y + 2 = x^{2} + y^{2} - 4 x + 2 y + 5$
$⟹ 2 x - 4 y - 3 = 0 \dots (1)$
$S A = S C$
$x^{2} + y^{2} - 2 x - 2 y + 2 = x^{2} + y^{2} - 6 x - 4 y + 11$
$⟹ 8 x + 2 y - 9 = 0 \dots (2)$
From $(1) (2)$
$⟹ y = \frac{- 3}{18} x = \frac{4}{3}$
Circum radius
$\sqrt{{(\frac{4}{3} - 1)}^{2} + {(\frac{- 3}{18} - 1)}^{2}}$

$\sqrt{0.11 + 1.36} = 1.2$

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