Question

find the coordinates of circumcentre of a triangle whose vertices are A(3,0) B(-1,-6)
AND C(4,-1) also find circum radius

Answer 1

$\huge{\bold{\ulcorner{\star\: Solution\: \star}\urcorner}}$

$\underline{Given \: \colon}$

Coordinate of triangle

A(3,0)

B(-1,-6)

C(4,-1)

$\underline{Solution\: \colon}$

Let the coordinates of the circumcentre of the triangle be P(x, y).Circumcentre of a triangle is equidistant from each of the vertices.

So PA = PB = PC

statement : circumcenter is equidistant from the vertices of triangle ABC.

So using distance formulae we are finding the coordinates of circumcenter P .

Now

$\sqrt{(x-x_1)^{2}+(y-y_1)^{2}} = \sqrt{(x-x_2)^{2} +(x-y_2)^{2}} = \sqrt{(x-x_3)^{2} + (y-y_3)^{2}}$

We have

$x_1 = 3 , y_1 = 0\\\\ x_2 = -1 , y_2= -6\\\\ x_3 = 4 , y_3 = -1$

Put the value in equation

Now

$\sqrt{(x-3)^{2}+(y-0)^{2}} = \sqrt{(x+1)^{2} +(x+6)^{2}} = \sqrt{(x-4)^{2} + (y+1)^{2}}$

$\sqrt{(x-3)^{2}+(y-0)^{2}} = \sqrt{(x+1)^{2} +(x+6)^{2}} \\\\ \cancel{x^{2}}+9-6x+\cancel{y^{2}}= \cancel{x^{2}}+1+2x+\cancel{y^{2}}+36+12y \\\\ 9-6x=1+2x+36+12y\\\\ 0 = (37-9)+(2x+6x)+12y\\\\ 0 = 28+8x+12y \\\\ 0= 4(7+2x+3y) \\\\ 2x+3y = -7$

Equation 1 : 2x+3y = -7


------------------------------------------------------------------------------------------------------------------


$\sqrt{(x+1)^{2} +(x+6)^{2}} = \sqrt{(x-4)^{2} + (y+1)^{2}} \\\\ \cancel{x^{2}}+\cancel{1}+2x+\cancel{y^{2}}+36+12y = \cancel{x^{2}} +16-8x+\cancel{y^{2}} +\cancel{1}+2y\\\\ 2x+36+12y = 16 -8x +2y \\\\ 2x+36+12y-16+8x-2y =0 \\\\ 10x+10y+20 =0 \\\\ x + y = -2$

Equation 2 : x + y = -2

Now in equation 2 : x = (-y -2) put this value in equation 1

$2x+3y\: = \: -7 \\\\ 2(-y-2)+3y = -7 \\\\ -2y-4 +3y = -7 \\\\ y-4 = -7 \\\\ y = -7+4 \\\\ y = -3\\\\ put\: the\: value \: of \: y \: in \: eq\: 2\\\\ x = (-y-2) \\\\ x = [-(-3)-2] \\\\ x = +3-2 \\\\ x = 1$

$\red{\underline{P(x,y) = (1\: , \:-3)}}$

------------------------------------------------------------------------------------------------------------------

$\bold{\underline{Circum - radius }}$

Take Coordinate A , B , C any one

Let I take B

B(-1,-6)
P( 1,-3 )

$\sqrt{(x-x_2)^{2} + (y-y_2)^{2}} \\\\\implies \sqrt{[1-(-1)]^{2} + [-3-(-6)]^{2}} \\\\\implies \sqrt{(2)^{2}+(3)^{2}} \\\\\implies \sqrt{4+9} \\\\\implies \sqrt{13} \:$

$PB \: = \: \sqrt{13} \: unit$

$PA = PB = PC =\sqrt{13}\: unit$

$\red{\underline{Circum-radius\: =\: \sqrt{13}}}$