The equation to the side "BC" of "`Delta ABC`" is "`x+5=0`" .If "`(-3,2)`" is the orthocentre and "(0,0)" is the circumcentre then radius of the circle is

In Delta ABC if the orthocentre is (1,2) and the circumcenter is (0,0) then centroid of Delta ABC is.

In a Delta ABC ,the equation of the side BC is ,2x-y=3 and its circumcentre and orthocentre are (3,5) and (2,3) respectively. If R is the circum radius of Delta ABC then [R^(2)] is equal to (where [.] denotes (G.I.F)

The equations of two sides of a triangle are 3x-2y+6=0 and 4x+5y-20 and the orthocentre is (1,1). Find the equation of the third side.

In Delta ABC orthocentre is (6,10) circumcentre is (2,3) and equation of side BC is 2x+y=17. Then the radius of the circumcircle is

In DeltaABC , equation of side BC is x+y-6=0, also the circumcentre and orhtocentre are (3, 1) and (2, 2) respectively, then the equation of the circumcircle of DeltaABC is

Let ABC be a triangle with equations of the sides AB,BC and CA respectively x-2=0,y-5=0 and 5x+2y-10=0 Then the orthocentre of the triangle lies on the line

In triangle ABC ,the equation of side BC is x-y=0. The circumcenter and orthocentre of triangle are (2,3) and (5,8), respectively. The equation of the circumcirle of the triangle is x^(2)+y^(2)-4x+6y-27=0x^(2)+y^(2)-4x-6y-27=0x^(2)+y^(2)+4x+6y-27=0x^(2)+y^(2)+4x+6y-27=0