The equation of the circle which has normals `x-1)x(y-2)=0` and a tangent `3x+4y=6` is `x^2+y^2-2x-4y+4=0` `x^2+y^2-2x-4y+5=0` `x^2+y^2=5` `(x-3)^2+9y-4)hat2=5`

The equation of the circle which has normals x-1)*(y-2)=0 and a tangent 3x+4y=6 is x^(2)+y^(2)-2x-4y+4=0x^(2)+y^(2)-2x-4y+5=0x^(2)+y^(2)=5(x-3)^(2)+(y-4)^(2)=5

The radius of the circle which has normals xy-2x-y+2=0 and a tangent 3x+4y-6=0 is

Circles x^(2)+y^(2)=4 and x^(2)+y^(2)-2x-4y+3=0

Suppose a x+b y+c=0 , where a ,ba n dc are in A P be normal to a family of circles. The equation of the circle of the family intersecting the circle x^2+y^2-4x-4y-1=0 orthogonally is (a)x^2+y^2-2x+4y-3=0 (b)x^2+y^2-2x+4y+3=0 (c)x^2+y^2+2x+4y+3=0 (d) x^2+y^2+2x-4y+3=0

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

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 equation of normal of x^2+y^2-2x+4y-5=0 at (2,1) is

The circles x^(2)+y^(2)-2x-4y+1=0 and x^(2)+y^(2)+4y-1 =0

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.