Question

The circumcentre of a triangle whose vertices are (-2, -3), (-1, 0), (7, -6)​

Answer 1

Let (x, y) be the circumcenter, which is equidistant from the vertices of the triangle (-2, -3), (-1, 0) and (7, -6) each.

Let the distance be d. So, by distance formula,

$\longrightarrow d^2=(x+2)^2+(y+3)^2$

$\longrightarrow d^2=x^2+y^2+4x+6y+13\quad\quad\dots(1)$

and,

$\longrightarrow d^2=(x+1)^2+(y-0)^2$

$\longrightarrow d^2=x^2+y^2+2x+1\quad\quad\dots(2)$

and,

$\longrightarrow d^2=(x-7)^2+(y+6)^2$

$\longrightarrow d^2=x^2+y^2-14x+12y+85\quad\quad\dots(3)$

Equating (1) and (2),

$\longrightarrow x^2+y^2+4x+6y+13=x^2+y^2+2x+1$

$\longrightarrow x=-3y-6\quad\quad\dots(4)$

Equating (2) and (3),

$\longrightarrow x^2+y^2+2x+1=x^2+y^2-14x+12y+85$

$\longrightarrow 16x-12y-84=0$

Putting value of  from (4),

$\longrightarrow 16(-3y-6)-12y-84=0$

$\longrightarrow -60y-180=0$

$\longrightarrow y=-3$

Then from (4),

$\longrightarrow x=-3(-3)-6$

$\longrightarrow x=3$

Hence the circumcenter of the triangle is (x, y) = (3, -3).

Answer 2

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hope it helps❤᭄࿐