In `Delta ABC` ,equation of BC is `x-y=0` if the circum center and orthocenter are`(2,3),(5,8)` then the circumradius is

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

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 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 equation to AB is 2x+3y-5=0 altitude through A is x-y+4=0 and altitude through B is 2x-y-1=0 Then the vertex C is

If combined equation of two sides AB and AC of a triangle is x^(2)-xy-2y^(2)+x+y=0 and (1,1) is orthocenter then circumradius of Delta ABC is 1) (5sqrt(2))/(3) 2) (10sqrt(2))/(3) 3) (3sqrt(2))/(5) 4) (3)/(10sqrt(2))

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 a triangle,ABC,the equation of the perpendicular bisector of AC is 3x-2y+8=0. If the coordinates of the points A and B are (1,-1)&(3,1) respectively,then the equation of the line BC & the centre of the circum-circle of the triangle ABC will be

If in Delta ABC , the distance of the vertices from the orthocenter are x,y, and z then prove that (a)/(x) + (b)/(y) + (c)/(z) = (abc)/(xyz)

The distance between circum centre and orthocenter of the triangle whose vertices are (0,0),(6,8) and (-4,3) is