If points `A and B` are (1, 0) and (0, 1), respectively, and point `C` is on the circle `x^2+y^2=1` , then the locus of the orthocentre of triangle `A B C` is `x^2+y^2=4` `x^2+y^2-x-y=0` `x^2+y^2-2x-2y+1=0` `x^2+y^2+2x-2y+1=0`

Tangents OA and OB are drawn from the origin to the circle (x-1)^2 + (y-1)^2 = 1 . Then the equation of the circumcircle of the triangle OAB is : (A) x^2 + y^2 + 2x + 2y = 0 (B) x^2 + y^2 + x + y = 0 (C) x^2 + y^2 - x - y = 0 (D) x^2+ y^2 - 2x - 2y = 0

A B C is a variable triangle such that A is (1, 2), and Ba n dC on the line y=x+lambda(lambda is a variable). Then the locus of the orthocentre of "triangle"A B C is x+y=0 (b) x-y=0 x^2+y^2=4 (d) x+y=3

From the points (3, 4), chords are drawn to the circle x^2+y^2-4x=0 . The locus of the midpoints of the chords is (a) x^2+y^2-5x-4y+6=0 (b) x^2+y^2+5x-4y+6=0 (c) x^2+y^2-5x+4y+6=0 (d) x^2+y^2-5x-4y-6=0

Prove that the centres of the circles x^2+y^2=1 , x^2+y^2+6x-2y-1=0 and x^2+y^2-12x+4y=1 are collinear

Prove that the centres of the circles x^2+y^2=1 , x^2+y^2+6x-2y-1=0 and x^2+y^2-12x+4y=1 are collinear

The locus of the centre of the circle cutting the circles x^2+y^2–2x-6y+1=0, x^2 + y^2 - 4x - 10y + 5 = 0 orthogonally is

The point of tangency of the circles x^2+ y^2 - 2x-4y = 0 and x^2 + y^2-8y -4 = 0 , is

The point of tangency of the circles x^2+ y^2 - 2x-4y = 0 and x^2 + y^2-8y -4 = 0 , is

ABC is a variable triangle such that A is (1, 2), and B and C on the line y=x+lambda(lambda is a variable). Then the locus of the orthocentre of DeltaA B C is (a) x+y=0 (b) x-y=0 x^2+y^2=4 (d) x+y=3

The equation of the circle passing through (1,2) and the points of intersection of the circles x^2+y^2-8x-6y+21=0 and x^2+y^2-2x-15=0 is