Answer

Verified

452.7k+ views

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.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Why Are Noble Gases NonReactive class 11 chemistry CBSE

Let X and Y be the sets of all positive divisors of class 11 maths CBSE

Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE

Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE

Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE

Trending doubts

Guru Purnima speech in English in 100 words class 7 english CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Which are the Top 10 Largest Countries of the World?

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Difference Between Plant Cell and Animal Cell

Change the following sentences into negative and interrogative class 10 english CBSE

Select the word that is correctly spelled a Twelveth class 10 english CBSE

Give 10 examples for herbs , shrubs , climbers , creepers