Question

Let d represent the distance from the center of the sphere x 2 + y 2 + z 2 − 6 x + 2 y = − 6 to the plane y = 2.

Find two integer values of A so that the center of a second sphere x 2 + y 2 + z 2 + A y + 6 z = 2 is also d units from the plane y = 2.
A = and (Give the smaller one first.)

Answer 1

`A=-10` and
`2`

Explanation:

The given sphere is
`x^2 + y^2 + z^2 -6x + 2y =-6`

This can be rearranged as
`(x-3)^2 -9+ (y+1)^2 -1 + (z-0)^2=-6`

`\Rightarrow (x-3)^2 + (y+1)^2 + (z-0)^2=4\\`

On comparing with the standard equation of a circle
`(x-x_0)^2 + (y-y_0)^2 + (z-z_0)^2=r^2` , with center
`(x_0,y_0,z_0)` and radius
`r`.

The center of the given circle is
`( 3, -1 ,0)`.

Now, as the perpendicular distance,
`d`, of a point
`(x_0,y_0,z_0)` from a plane
`ax+by+cz=p` is

`d= \left | \frac {xx_0+yy_0+zz_0-p}{\sqrt {a^2+b^2 +c^2}} \right|`.

`\Rightarrow d= \left | \frac {0* 3+1* (-1)+0-2}{\sqrt {1^2}} \right|` [ as the given plane is y=2]

`\Rightarrow d=3 \; \cdots (i)`

Similarly, the other equation of the circle,
`x^2 + y^2 + z^2 + Ay + 6z = 2,` can be rearranged as
`(x-0)^2 + (y+\frac {A}{2})^2 + (z+3)^2=11+\frac {A^2}{4}.`

Here, the center for this circle is
`\left( 0, -\frac {A}{2}, -3\rigfht).`

From the given condition, the distance of the center from the plane
`y=2` is
`d`.

`\Rightarrow \left | \frac {0+1 * \left (-\frac {A}{2}\right)+0* (-3)-2}{\sqrt {1^2}} \right|=3` [from equation (i) ]

`\Rightarrow \left | -\frac {A}{2}-2\right|=3`

`\Rightarrow \frac {A}{2}+2=\pm 3`

`\Rightarrow \frac {A}{2}=-2\pm 3`

`\Rightarrow A=2* (-2-3)` and
`2* (-2+3)`

`\Rightarrow A=-10` and
`2`.