$A$ circle passes through $(0, 0)$ and $(1, 0)$ and touches the circle ${x^2} + {y^2} = 9$. Find the centre of the circle.

Let $C_1$ be the circle of radius $1$ with center at the origin. Let $C_2$ be the circle of radius $r$ with center at the point $A=(4,1)$,where $1 < r < 3$. Two distinct common tangents $PQ$ and $ST$ of $C_1$ and $C_2$ are drawn. The tangent $PQ$ touches $C_1$ at $P$ and $C_2$ at $Q$. The tangent $ST$ touches $C_1$ at $S$ and $C_2$ at $T$. Midpoints of the line segments $PQ$ and $ST$ are joined to form a line which meets the $x$-axis at a point $B$. If $AB=\sqrt{5}$,then the value of $r^2$ is

The equation of the circle which cuts the circles $x^2+y^2+4x-7=0$,$2x^2+2y^2+3x+5y-9=0$,and $x^2+y^2+y=0$ orthogonally is

The equation of the circle which passes through the point $(1, 1)$ and intersects the given circles $x^2 + y^2 + 2x + 4y + 6 = 0$ and $x^2 + y^2 + 4x + 6y + 2 = 0$ orthogonally is:

If the lengths of tangents drawn from a point $P$ to the circles $x^2+y^2-8x+40=0$,$5x^2+5y^2-25x+80=0$,and $x^2+y^2-8x+16y+160=0$ are equal,then the point $P$ 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