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.

Prove that the centres of the three circles x^(2)+y^(2)-4x6y12=0,x^(2)+y^(2)+2x+4y-5=0 and x^(2)+y^(2)-10x16y+7=0 are collinear.

Find the centre and radius of the circle x^(2) + y^(2) - 6x + 4y - 12 =0 .

Two circles x^(2) + y^(2) - 4x + 10y + 20 = 0 and x^(2) + y^(2) + 8x - 6y - 24= 0

Centres of the three circles x^2+y^2 - 4x - 6y - 14=0, x^2 +y^2-4x-6y-14=0 and x^2+y^2-10x-16y+7=0(A) are the vertices of a right triangle (B) the vertices of an isosceles triangle which is not regular (C) vertices of a regular triangle (D) are collinear ts inht lines, at right angled to each other.

If the two circle x^(2) + y^(2) - 10 x - 14y + k = 0 and x^(2) + y^(2) - 4x - 6y + 4 = 0 are orthogonal , find k.

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

Equation of radical axis of the circles x^(2) + y^(2) - 3x - 4y + 5 = 0 and 2x^(2) + 2y^(2) - 10x - 12y + 12 = 0 is

The locus of the centers of the circles which cut the circles x^(2) + y^(2) + 4x – 6y + 9 = 0 and x^(2) + y^(2) – 5x + 4y – 2 = 0 orthogonally is

y-1=m_1(x-3) and y - 3 = m_2(x - 1) are two family of straight lines, at right angled to each other. The locus of their point of intersection is: (A) x^2 + y^2 - 2x - 6y + 10 = 0 (B) x^2 + y^2 - 4x - 4y +6 = 0 (C) x^2 + y^2 - 2x - 6y + 6 = 0 (D) x^2 + y^2 - 4x - by - 6 = 0