Three sides of a triangle are represented by lines whose combined equation is `(2x+y-4) (xy-4x-2y+8) = 0`, then the equation of its circumcircle will be : (A) `x^2 + y^2 - 2x - 4y = 0` (B) `x^2 + y^2 + 2x + 4y = 0` (C) `x^2 + y^2 - 2x + 4y = 0` (D) `x^2 + y^2 + 2x - 4y = 0`

A triangle is formed by the lines whose combined equation is given by (x-y-4)(xy-2x-y+2)=0 The equation of its circumcircle is:

The triangle formed by the lines whose combined equation is (y^2 - 4xy - x^2)(x + y - 1) = 0 is

Find the lines whose combined equation is 6x^2+5x y-4y^2+7x+13 y-3=0

Find the lines whose combined equation is 6x^2+5x y-4y^2+7x+13 y-3=0

Orthocentre of the triangle formed by the lines whose combined equation is (y ^(2) - 6xy -x ^(2)) ( 4x -y +7) =0 is

The three lines whose combined equation is y^(3)-4x^(2)y=0 form a triangle which is

The equation of the circumcircle of the triangle formed by the lines whose combined equation is given by (x+y-4) (xy-2x-y+2)=0, is

A line intersects x-axis at A(2, 0) and y-axis at B(0, 4) . A variable lines PQ which is perpendicular to AB intersects x-axis at P and y-axis at Q . AQ and BP intersect at R . Locus of R is : (A) x^2 + y^2 - 2x + 4y = 0 (B) x^2 + y^2 + 2x + 4y = 0 (C) x^2 + y^2 - 2x - 4y=0 (D) x^2 + y^2 + 2x - 4y = 0

Prove that the centres of the three circles x^2 + y^2 - 4x – 6y – 12 = 0,x^2+y^2 + 2x + 4y -5 = 0 and x^2 + y^2 - 10x – 16y +7 = 0 are collinear.

Find the equation of circumcircle of the triangle whose sides are 2x +y = 4,x + y = 6 and x +2y=5 .