Answer
Verified
480.3k+ 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
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE
Which are the Top 10 Largest Countries of the World?
How do you graph the function fx 4x class 9 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Why is there a time difference of about 5 hours between class 10 social science CBSE