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Hint: From circumcenter to vertices of triangle are equidistant.So,we have to equate the distances from circumcentre to vertices of triangle.

Let $$A( - 2, - 3),B( - 1,0),C( - 7, - 6)$$ be the vertices of the triangle and $P(x,y)$ be the required circumcenter. Also, $P{A^2} = P{B^2} = P{C^2}$

Formula for finding distance between two points$\left( {{x_1},{y_1}} \right),\left( {{x_2},{y_2}} \right)$ $ = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $

Distance between Points P and A is

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

$ \Rightarrow P{A^2} = {(x + 2)^2} + {(y + 1)^2}$

$ \Rightarrow P{A^2} = {x^2} + 4 + 4x + {y^2} + 9 + 6y$ .... (1)

Distance between Points P and B is

$PB = \sqrt {{{(x - ( - 1))}^2} + {{(y - 0)}^2}} $

$ \Rightarrow P{B^2} = {(x + 1)^2} + {y^2}$

$ \Rightarrow P{B^2} = {x^2} + 1 + 2x + {y^2}$ .... (2)

Distance between Points P and C is

$PC = \sqrt {{{(x - 7)}^2} + {{(y - ( - 6))}^2}} $

$ \Rightarrow P{C^2} = {(x - 7)^2} + {(y + 6)^2}$

$ \Rightarrow P{C^2} = {x^2} + 49 - 14x + {y^2} + 36 + 12y$.... (3)

From equations (1) and (2) we make PA = PB

${x^2} + 4x + {y^2} + 6y + 13 = {x^2} + 2x + {y^2} + 1$

$ \Rightarrow 4x + 6y + 13 = 1 + 2x$

$ \Rightarrow 2x + 6y + 12 = 0$

$ \Rightarrow x + 3y + 6 = 0$ ... (4)

From equations (2) and (3) we make PB = PC

${x^2} + 1 + 2x + {y^2} = {x^2} - 14x + {y^2} + 12y + 85$

$ \Rightarrow 2x - 12y + 14x = 84$

$ \Rightarrow 16x - 12y = 84$

$ \Rightarrow 4x - 3y = 21$

From equation (4) we can put $x = - 3y - 6$

$ \Rightarrow 4( - 3y - 6) - 3y = 21$

$\eqalign{

& \Rightarrow - 12y - 24 - 3y = 21 \cr

& \Rightarrow 15y = - 45 \cr} $

$ \Rightarrow y = \frac{{ - 45}}{{15}} = - 3$

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

We got x, y values which are coordinates of the circumference.

$\therefore $The vertices of circumcenter = (3, -3)

Note: The circumcenter of a triangle is a point where the perpendicular bisector of the sides of that particular triangle. That means the circumcenter is equidistant from each vertex of the triangle. It can be inside or outside of the triangle based on the type of triangle.

Let $$A( - 2, - 3),B( - 1,0),C( - 7, - 6)$$ be the vertices of the triangle and $P(x,y)$ be the required circumcenter. Also, $P{A^2} = P{B^2} = P{C^2}$

Formula for finding distance between two points$\left( {{x_1},{y_1}} \right),\left( {{x_2},{y_2}} \right)$ $ = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $

Distance between Points P and A is

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

$ \Rightarrow P{A^2} = {(x + 2)^2} + {(y + 1)^2}$

$ \Rightarrow P{A^2} = {x^2} + 4 + 4x + {y^2} + 9 + 6y$ .... (1)

Distance between Points P and B is

$PB = \sqrt {{{(x - ( - 1))}^2} + {{(y - 0)}^2}} $

$ \Rightarrow P{B^2} = {(x + 1)^2} + {y^2}$

$ \Rightarrow P{B^2} = {x^2} + 1 + 2x + {y^2}$ .... (2)

Distance between Points P and C is

$PC = \sqrt {{{(x - 7)}^2} + {{(y - ( - 6))}^2}} $

$ \Rightarrow P{C^2} = {(x - 7)^2} + {(y + 6)^2}$

$ \Rightarrow P{C^2} = {x^2} + 49 - 14x + {y^2} + 36 + 12y$.... (3)

From equations (1) and (2) we make PA = PB

${x^2} + 4x + {y^2} + 6y + 13 = {x^2} + 2x + {y^2} + 1$

$ \Rightarrow 4x + 6y + 13 = 1 + 2x$

$ \Rightarrow 2x + 6y + 12 = 0$

$ \Rightarrow x + 3y + 6 = 0$ ... (4)

From equations (2) and (3) we make PB = PC

${x^2} + 1 + 2x + {y^2} = {x^2} - 14x + {y^2} + 12y + 85$

$ \Rightarrow 2x - 12y + 14x = 84$

$ \Rightarrow 16x - 12y = 84$

$ \Rightarrow 4x - 3y = 21$

From equation (4) we can put $x = - 3y - 6$

$ \Rightarrow 4( - 3y - 6) - 3y = 21$

$\eqalign{

& \Rightarrow - 12y - 24 - 3y = 21 \cr

& \Rightarrow 15y = - 45 \cr} $

$ \Rightarrow y = \frac{{ - 45}}{{15}} = - 3$

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

We got x, y values which are coordinates of the circumference.

$\therefore $The vertices of circumcenter = (3, -3)

Note: The circumcenter of a triangle is a point where the perpendicular bisector of the sides of that particular triangle. That means the circumcenter is equidistant from each vertex of the triangle. It can be inside or outside of the triangle based on the type of triangle.

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