If the lines $l_1x + m_1y + n_1 = 0$ and $l_2x + m_2y + n_2 = 0$ cut the axes at concyclic points,then

The centre of the circle $r^2-4r(\cos \theta+\sin \theta)-4=0$ in Cartesian coordinates is

Let the circle $S: 36 x^{2}+36 y^{2}-108 x+120 y+C=0$ be such that it neither intersects nor touches the coordinate axes. If the point of intersection of the lines $x-2 y=4$ and $2 x-y=5$ lies inside the circle $S$,then :

The equation of the circle whose centre lies on the lines $3x - y - 4 = 0$ and $x + 3y + 2 = 0$ and has an area of $154$ square units is:

The equation of a circle touching the coordinate axes and the line $x \cos \alpha + y \sin \alpha = 2$ can be:

$A$ circle cuts off positive intercepts $5$ and $6$ on the $x$ and $y$ axes respectively,and passes through the origin. Then the equation of the circle is