The 3-D Coordinate System — Question 10

Question 10 – Solution

Strategy The center lies on the perpendicular-bisector plane of the two axial intersection points. Tangency to the $xy$-plane relates the radius to the center’s height.

Center coordinates The midpoint of the two intersections has $x$-coordinate $1$, so write the center as $$C=(1,-2,k).$$ The squared radius, using $(5,0,0)$, is $$r^2=(5-1)^2+(0+2)^2+k^2=20+k^2.$$ Tangency to the $xy$-plane requires $r=|k|$, hence $r^2=k^2$. Combining these equations would give $k^2=20+k^2$, an impossibility.

Conclusion Therefore .

Consistency review The two axial intersection points already force every possible center onto $x=1$. Requiring $y=-2$ means the center has a nonzero horizontal offset from the $x$-axis. Consequently the radius needed to reach either axial point is strictly greater than $|k|$, while tangency to the $xy$-plane requires it to equal $|k|$. The contradiction is geometric, not an algebraic accident.