Obtain the circumcentre of $\Delta ABC$ with vertices $A (-3, -1)$,$B (-1, 3)$ and $C (6, 2)$.
(P(2, -1)) Let $P(x, y)$ be the circumcentre of $\Delta ABC$.
Since $P$ is the circumcentre,the distance from $P$ to each vertex is equal,i.e.,$PA = PB = PC$,which implies $PA^2 = PB^2 = PC^2$.
$1$. $PA^2 = (x + 3)^2 + (y + 1)^2 = x^2 + 6x + 9 + y^2 + 2y + 1 = x^2 + y^2 + 6x + 2y + 10$
$2$. $PB^2 = (x + 1)^2 + (y - 3)^2 = x^2 + 2x + 1 + y^2 - 6y + 9 = x^2 + y^2 + 2x - 6y + 10$
$3$. $PC^2 = (x - 6)^2 + (y - 2)^2 = x^2 - 12x + 36 + y^2 - 4y + 4 = x^2 + y^2 - 12x - 4y + 40$
Equating $PA^2 = PB^2$:
$x^2 + y^2 + 6x + 2y + 10 = x^2 + y^2 + 2x - 6y + 10$
$4x + 8y = 0 \implies x = -2y$ ... $(1)$
Equating $PB^2 = PC^2$:
$x^2 + y^2 + 2x - 6y + 10 = x^2 + y^2 - 12x - 4y + 40$
$14x - 2y = 30 \implies 7x - y = 15$ ... $(2)$
Substitute $(1)$ into $(2)$:
$7(-2y) - y = 15$
$-14y - y = 15 \implies -15y = 15 \implies y = -1$
Using $y = -1$ in $(1)$:
$x = -2(-1) = 2$
Thus,the circumcentre is $P(2, -1)$.
Since $P$ is the circumcentre,the distance from $P$ to each vertex is equal,i.e.,$PA = PB = PC$,which implies $PA^2 = PB^2 = PC^2$.
$1$. $PA^2 = (x + 3)^2 + (y + 1)^2 = x^2 + 6x + 9 + y^2 + 2y + 1 = x^2 + y^2 + 6x + 2y + 10$
$2$. $PB^2 = (x + 1)^2 + (y - 3)^2 = x^2 + 2x + 1 + y^2 - 6y + 9 = x^2 + y^2 + 2x - 6y + 10$
$3$. $PC^2 = (x - 6)^2 + (y - 2)^2 = x^2 - 12x + 36 + y^2 - 4y + 4 = x^2 + y^2 - 12x - 4y + 40$
Equating $PA^2 = PB^2$:
$x^2 + y^2 + 6x + 2y + 10 = x^2 + y^2 + 2x - 6y + 10$
$4x + 8y = 0 \implies x = -2y$ ... $(1)$
Equating $PB^2 = PC^2$:
$x^2 + y^2 + 2x - 6y + 10 = x^2 + y^2 - 12x - 4y + 40$
$14x - 2y = 30 \implies 7x - y = 15$ ... $(2)$
Substitute $(1)$ into $(2)$:
$7(-2y) - y = 15$
$-14y - y = 15 \implies -15y = 15 \implies y = -1$
Using $y = -1$ in $(1)$:
$x = -2(-1) = 2$
Thus,the circumcentre is $P(2, -1)$.
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