Find the coordinates of the circumcentre of the triangle whose vertices are (3,0) , (-1,-6) and (4,-1) Also - Maths - Coordinate Geometry

Question

Find the coordinates of the circumcentre of the triangle whose
vertices are (3,0) , (-1,-6) and (4,-1) Also find its
circumradius

Manish Bharti · Accepted Answer

Let the coordinates of circumcentre O be (x,y)
Distance of circumcentre from the vertices A(3,0), B(-1,-6) and C(4,-1) of the triangle will be equal as it is equal to the radius of the circle.
Therefore,
OA = OB
$\sqrt{(x - 3)^{2} + (y - 0)^{2}} = \sqrt{(x + 1)^{2} + (y + 6)^{2}} (x - 3)^{2} + y^{2} = (x + 1)^{2} + {(y + 6)}^{2} x^{2} + 9 - 6 x + y^{2} = x^{2} + 1 + 2 x + y^{2} + 12 y + 36 - 8 x - 12 y - 28 = 0 2 x + 3 y + 7 = 0 - - - - (1)$
and OB = OC
$\sqrt{(x + 1)^{2} + (y + 6)^{2}} = \sqrt{(x - 4)^{2} + (y + 1)^{2}} (x + 1)^{2} + {(y + 6)}^{2} = (x - 4)^{2} + {(y + 1)}^{2} x^{2} + 1 + 2 x + y^{2} + 12 y + 36 = x^{2} + 16 - 8 x + y^{2} + 1 + 2 y 10 x + 10 y + 20 = 0 x + y + 2 = 0 - - - - (2)$
By subtracting 2 $\times$ equation (2) from Equation (1), we get
$y + 3 = 0 \Rightarrow y = - 3$
Substituting the value of y in equation (2) ,we get
$x - 3 + 2 = 0 \Rightarrow x = 1$
therefore, coordinates of circumcentre are $(1, - 3)$

and Radius of circles is
$O A = \sqrt{(1 - 3)^{2} + (- 3 - 0)^{2}} = \sqrt{13}$

Find the coordinates of the circumcentre of the triangle whose vertices are (3,0) , (-1,-6) and (4,-1) . Also find its circumradius.