Respuesta :
Answer:
[tex]A=-10[/tex] and [tex]2[/tex]
Step-by-step explanation:
The given sphere is [tex]x^2 + y^2 + z^2 -6x + 2y =-6[/tex]
This can be rearranged as [tex](x-3)^2 -9+ (y+1)^2 -1 + (z-0)^2=-6[/tex]
[tex]\Rightarrow (x-3)^2 + (y+1)^2 + (z-0)^2=4\\[/tex]
On comparing with the standard equation of a circle [tex](x-x_0)^2 + (y-y_0)^2 + (z-z_0)^2=r^2[/tex] , with center [tex](x_0,y_0,z_0)[/tex] and radius [tex]r[/tex].
The center of the given circle is [tex]( 3, -1 ,0)[/tex].
Now, as the perpendicular distance, [tex]d[/tex], of a point [tex](x_0,y_0,z_0)[/tex] from a plane [tex]ax+by+cz=p[/tex] is
[tex]d= \left | \frac {xx_0+yy_0+zz_0-p}{\sqrt {a^2+b^2 +c^2}} \right|[/tex].
[tex]\Rightarrow d= \left | \frac {0\times 3+1\times (-1)+0-2}{\sqrt {1^2}} \right|[/tex] [ as the given plane is y=2]
[tex]\Rightarrow d=3 \; \cdots (i)[/tex]
Similarly, the other equation of the circle, [tex]x^2 + y^2 + z^2 + Ay + 6z = 2,[/tex] can be rearranged as [tex](x-0)^2 + (y+\frac {A}{2})^2 + (z+3)^2=11+\frac {A^2}{4}.[/tex]
Here, the center for this circle is [tex]\left( 0, -\frac {A}{2}, -3\rigfht).[/tex]
From the given condition, the distance of the center from the plane [tex]y=2[/tex] is [tex]d[/tex].
[tex]\Rightarrow \left | \frac {0+1 \times \left (-\frac {A}{2}\right)+0\times (-3)-2}{\sqrt {1^2}} \right|=3[/tex] [from equation (i) ]
[tex]\Rightarrow \left | -\frac {A}{2}-2\right|=3[/tex]
[tex]\Rightarrow \frac {A}{2}+2=\pm 3[/tex]
[tex]\Rightarrow \frac {A}{2}=-2\pm 3[/tex]
[tex]\Rightarrow A=2\times (-2-3)[/tex] and [tex]2\times (-2+3)[/tex]
[tex]\Rightarrow A=-10[/tex] and [tex]2[/tex].
