Sphere

A variable sphere of constant radius passes through the point $\left(0, 0, 1\right)$ and touches the $xy$-plane. The locus of the centre of the sphere is

For $x=\left(x_1,x_2,x_3\right)\in {\mathrm{ℝ}}^3$, define $\left|x\right|=\sqrt{x_1^2+x_2^2+x_3^2}$. For $a=\left(1,4,4\right)$ and $b=$$\left(1,0,1\right)$, the maximum value of $\left|x\right|$ satisfying $\left|x-a\right|=2\left|x-b\right|$ is

A spherical ball is kept at the corner of a rectangular room such that the ball touches two (perpendicular) walls and lies on the floor. If a point on the sphere is at distance of $9,16,25$ from the two walls and the floor, then a possible radius of the sphere is

The shortest distance from the origin to a variable point on the sphere ${\left(x-2\right)}^2+{\left(y-3\right)}^2+{\left(z-6\right)}^2=1$

The radius of the circle in which the sphere $x^2+y^2+z^2+2x-2y-4z-19=0$ is cut by the plane $x+2y+2z+7=0$ is

The shortest distance from the plane $12x+4y+3z=327$ to the sphere $x^2+y^2+z^2+4x-2y-6z=155$ is

The radius of sphere $x+2y+2z=15$ and $x^2+y^2+z^2-2y-4z=11$ is

The plane $x+2y-z=4$ cuts the sphere $x^2+y^2+z^2$ $-x+z-2=0$ in a circle of radius

If two spheres of radii $r_1$ and $r_2$ cut orthogonally, then the radius of the common circle is

Co-ordinate of a point equidistant from the points $\left(\mathrm{0,0},0\right),\mathrm{ }\left(a,\mathrm{ }0,\mathrm{ }0\right),\mathrm{ }\left(0,\mathrm{ }b,\mathrm{ }0\right),\mathrm{ }\left(0,\mathrm{ }0,\mathrm{ }c\right)$ is

If a sphere of radius $r$ passes through the origin, then the extremities of the diameter parallel to $x$-axis lie on each of the spheres

The centre of sphere passes through four points $\left(0, 0, 0\right), \left(0, 2, 0\right), \left(1, 0, 0\right) \& \left(0, 0, 4\right)$ is

Which of the following is an equation of a sphere?

The plane of intersection of spheres $x^2+y^2+{\mathcal{z}}^2+2x+2y+2\mathcal{z}=2$ and

$2x^2+2y^2+2{\mathcal{z}}^2+4x+2y+4\mathcal{z}=0$ is

The equation of sphere which passes through the sphere $x^2+y^2+{\mathcal{z}}^2=9$, the plane $2x+3y+4\mathcal{z}=5$ and point (1, 2, 3) is

The shortest distance from the point $(1, 2, -1)$ to the surface of the sphere $x^2+y^2+{\mathcal{z}}^2=24$ is

Let $\left(3, 4,-1\right)$ and $\left(-1, 2,3\right)$ are the end points of a diameter of sphere. Then, the radius of the sphere is equal to

The smallest radius of the sphere passing through (1, 0, 0), (0, 1, 0) and (0, 0, 1) is

The radius of the sphere $x^2+y^2+z^2=x+2y+3z=0$ is, 

If the plane $x+y+z=-a\sqrt{3}$ touches the sphere , ${2x}^2+2y^2+2z^2-2x+4y-4z+3=0$, then the value of $a$ is,