The 3-D Coordinate System — Question 7

Question 7

Find every sphere satisfying all of the following conditions:

1. The sphere lies entirely in the octant $$x\ge 0,\qquad y\le 0,\qquad z\ge 0.$$

2. It is tangent to each of the three coordinate planes.

3. It passes through the point $$P=(1,-2,3).$$

For each sphere, state its center, radius, and Cartesian equation. Then give a geometric explanation for why two different spheres satisfy the same conditions.

Question 7 – Solution

Strategy Tangency to all three coordinate planes makes the center’s coordinate magnitudes equal to the radius. The specified octant fixes their signs.

Center condition If the radius is $r>0$, then the center must be $$C=(r,-r,r).$$ Since $P$ lies on the sphere, $$(1-r)^2+(-2+r)^2+(3-r)^2=r^2.$$ Expanding gives $2r^2-12r+14=0$, or $$r^2-6r+7=0.$$ Hence $r=3\pm\sqrt 2$, both positive.

Results The two spheres are $$\boxed{(x-r)^2+(y+r)^2+(z-r)^2=r^2,\qquad r=3\pm\sqrt 2}.$$

Why two? Along the line of possible centers $(r,-r,r)$, the distance to $P$ first decreases and then increases, so the equation “distance to center equals $r$” can be met at two scales. Direct substitution verifies both roots, and each sphere has extreme coordinates touching—but not crossing—the three coordinate planes.