The equation of the circle inscribed in a square formed by the lines $x+y-2=0$,$x+y-6=0$,$x-y+1=0$ and $x-y+5=0$ is

Find the equation of a circle that touches the $Y$-axis at a distance of $4$ units from the origin and cuts an intercept of $6$ units on the $X$-axis.

For each natural number $k$,let $C_k$ denote the circle with radius $k$ centimeters and center at the origin. On the circle $C_k$,a particle moves $k$ centimeters in the counter-clockwise direction. After completing its motion on $C_k$,the particle moves to $C_{k+1}$ in the radial direction. The motion of the particle continues in this manner. The particle starts at $(1, 0)$. If the particle crosses the positive direction of the $x$-axis for the first time on the circle $C_n$,then $n$ is equal to

The equation of the circle which touches the circle $x^2+y^2-6x+6y+17=0$ externally and has the lines $x^2-3xy-3x+9y=0$ as two normals,is

$A$ square is inscribed in the circle $x^2 + y^2 - 6x + 8y - 103 = 0$ with its sides parallel to the coordinate axes. Then the distance of the vertex of the square which is nearest to the origin is

The total number of common tangents of $x^{2}+y^{2}-6x-8y+9=0$ and $x^{2}+y^{2}=1$ is