If the centre of a circle which passing through the points of intersection of the circles ${x^2} + {y^2} - 6x + 2y + 4 = 0$and ${x^2} + {y^2} + 2x - 4y - 6 = 0$ is on the line $y = x$, then the equation of the circle is

Let $C_i \equiv  x^2 + y^2 = i^2 (i = 1,2,3)$ are three circles. If there are $4i$ points on circumference of circle $C_i$. If no three of all the points on three circles are collinear then number of triangles which can be formed using these points whose circumcentre does not lie on origin, is-

The number of circles touching the line $y - x = 0$ and the $y$-axis is

$P$ is a point $(a, b)$ in the first quadrant. If the two circles which pass through $P$ and touch both the co-ordinate axes cut at right angles, then :

The locus of centre of the circle which cuts the circles${x^2} + {y^2} + 2{g_1}x + 2{f_1}y + {c_1} = 0$ and ${x^2} + {y^2} + 2{g_2}x + 2{f_2}y + {c_2} = 0$ orthogonally is

The circle on the chord $x\cos \alpha + y\sin \alpha = p$ of the circle ${x^2} + {y^2} = {a^2}$ as diameter has the equation