The equation of the line joining the centres of the circles belonging to the coaxial system of circles $4x^2 + 4y^2 - 12x + 6y - 3 + \lambda(x + 2y - 6) = 0$ is

The equation of the circle described on the chord $3x + y + 5 = 0$ of the circle $x^2 + y^2 = 16$ as diameter is:

The equation of the circle passing through $(0,0)$ and cutting orthogonally the circles $x^2+y^2+6x-15=0$ and $x^2+y^2-8y-10=0$ is

If the radical centre of the three circles $x^2+y^2=1$,$x^2+y^2-2x-3=0$,and $x^2+y^2-2y-3=0$ is $C(\alpha, \beta)$ and $r$ is the sum of the radii of the given circles,then the equation of the circle with $C(\alpha, \beta)$ as centre and $r$ as radius is:

If a circle passing through the point $(1,1)$ cuts the circles $x^2+y^2+4x-5=0$ and $x^2+y^2-4y+3=0$ orthogonally,then the center of that circle is:

$A$ circle $C$ passing through the point $(1, 1)$ bisects the circumference of the circle $x^2+y^2-2x=0$. If $C$ is orthogonal to the circle $x^2+y^2+2y-3=0$,then the centre of the circle $C$ is