The equation of the circle which cuts the three circles `x^2+y^2-4x-6y+4=0, x^2+y^2-2x-8y+4=0, x^2+y^2-6x-6y+4=0` orthogonally is

The equation of the circle which cuts the three circles x^(2)+y^(2)-4x-6y+4=0 , x^(2)+y^(2)-2x-8y+4=0 and x^(2)+y^(2)-6x-6y+4=0 orthogonally is

Find the equation of the circle which cuts the three circles x^(2)+y^(2)-3x-6y+14=0,x^(2)+y^(2)-x-4y+8=0 and x^(2)+y^(2)+2x-6y+9=0 orthogonally.

The equation of a circle which cuts the three circles x^(2)+y^(2)+2x+4y+1=0,x^(2)+y^(2)-x-4y+8=0 and x^(2)+y^(2)+2x-6y+9=0 orthogonlly is

The centre of circle cutting the three circles x^(2)+y^(2)-4x-4y+5=0 , x^(2)+y^(2)-4x-6y+5=0 , x^(2)+y^(2)-4x-8y+5=0 orthogonally is

The equation of the normal at (1,2) to the circle x^(2)-y^(2)-4x-6y-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

Find the equation of the circle which cuts each of the circles x^(2)+y^(2)=4x^(2)+y^(2)-6x-8y.+10=0&x^(2)+y^(2)+2x-4y-2=0 at the extremities of a diameter