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.

Answer 1

Answer:

$A=-10$ and  $2$

Step-by-step 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

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

$\Rightarrow d= \left | \frac {0\times 3+1\times (-1)+0-2}{\sqrt {1^2}} \right|$ [ 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 $d$.

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

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

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

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

$\Rightarrow A=2\times (-2-3)$ and $2\times (-2+3)$

$\Rightarrow A=-10$ and  $2$.