Answer:
See explanation
Step-by-step explanation:
The equation of the shpere canterd at point [tex](a,b,c)[/tex] with radius [tex]R[/tex] is
[tex](x-a)^2+(y-b)^2+(z-c)^2=R^2[/tex]
The center of the sphere is (2,-9,1), the radius is 2, then the equation is
[tex](x-2)^2+(y+9)^2+(z-1)^2=2^2\\ \\(x-2)^2+(y+9)^2+(z-1)^2=4[/tex]
Now, find the intersection of this sphere with the plane z=2. Substitute 2 instead of z into the sphere equation:
[tex](x-2)^2 +(y+9)^2+(2-1)^2=4\\ \\(x-2)^2+(y+9)^2+1=4\\ \\(x-2)^2+(y+9)^2=3[/tex]
This is the equation of the circle lying in the plane z=2 with center at (2,-9,2) and radius [tex]\sqrt{3}.[/tex]
