The two circles $x^2 + y^2 - 2x + 22y + 5 = 0$ and $x^2 + y^2 + 14x + 6y + k = 0$ intersect orthogonally provided $k$ is equal to

One of the limit points of the coaxial system of circles containing $x^2 + y^2 - 6x - 6y + 4 = 0$ and $x^2 + y^2 - 2x - 4y + 3 = 0$ is:

The equation of the circle which passes through the intersection of ${x^2} + {y^2} + 13x - 3y = 0$ and $2{x^2} + 2{y^2} + 4x - 7y - 25 = 0$ and whose centre lies on $13x + 30y = 0$ is

The centre of the circle passing through the point $(1,1)$ and orthogonal to the circles $x^2+y^2+3x-5y+7=0$ and $x^2+y^2-6x-10y+9=0$ is

The line $Ax + By + C = 0$ cuts the circle $x^2 + y^2 + ax + by + c = 0$ at points $P$ and $Q$,and the line $A'x + B'y + C' = 0$ cuts the circle $x^2 + y^2 + a'x + b'y + c' = 0$ at points $R$ and $S$. If the four points $P, Q, R,$ and $S$ are concyclic,then $D = \left| {\begin{array}{*{20}{c}}{a - a'}&{b - b'}&{c - c'}\\A&B&C\\{A'}&{B'}&{C'}\end{array}} \right| = $

Let $x-4=0$ be the radical axis of two circles which are intersecting orthogonally. If $x^2+y^2=36$ is one of those circles,then the other circle is