A circle passes through the origin and has its centre on $y = x$. If it cuts ${x^2} + {y^2} - 4x - 6y + 10 = 0$ orthogonally, then the equation of the circle is

Let $C: x^2+y^2=4$ and $C^{\prime}: x^2+y^2-4 \lambda x+9=0$ be two circles. If the set of all values of $\lambda$ so that the circles $\mathrm{C}$ and $\mathrm{C}^{\prime}$ intersect at two distinct points, is ${R}-[a, b]$, then the point $(8 a+12,16 b-20)$ lies on the curve:

Let $S = 0$ is the locus of centre of a variable circle which intersect the circle $x^2 + y^2 -4x -6y = 0$ orthogonally at $(4, 6)$ . If $P$ is a variable point of $S = 0$ , then least value of $OP$ is (where $O$ is origin)

Consider a circle $C_1: x^2+y^2-4 x-2 y=\alpha-5$.Let its mirror image in the line $y=2 x+1$ be another circle $C _2: 5 x ^2+5 y ^2-10 fx -10 gy +36=0$.Let $r$ be the radius of $C _2$. Then $\alpha+ r$ is equal to $......$.

For the two circles $x^2 + y^2 = 16$ and $x^2 + y^2 -2y = 0,$ there is/are

The equation of the circle having the lines ${x^2} + 2xy + 3x + 6y = 0$ as its normals and having size just sufficient to contain the circle $x(x - 4) + y(y - 3) = 0$is