Find the equation of the sphere which touches the three coordinate planes and the plane 2x+y+2z=6. Please help me to solve this problem. What does three coordinate planes mean?
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$\forall x (x, 0, 0) $ etc – Emil Nov 01 '16 at 06:29
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The touching means tangent. The sphere is pushed to the corner. Assume it is in the positive section, than the center of it is on a line that is between the three axis. Now you may expand it till it touches the plane - and actually there are two solutions. – Moti Nov 01 '16 at 06:41
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@Moti You gave the perfect image, although I feel that there should be one solution in the positive section. To the MO : the three coordinate planes have equations (1) x=0 (2) y=0 (3) z=0. – Duchamp Gérard H. E. Nov 01 '16 at 06:55
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A sphere that touches all the coordinate planes has center of the form $(\pm r, \pm r, \pm r)$. If the sphere lies in the first octant, then we have $$\left|\frac{2r+r+2r-6}{3}\right| = r$$ and hence $$4r^2 - 15r + 9 = 0$$ or $r = 3/4$ or $r=3$. Thus the spheres are $$(x-3)^2 + (y-3)^2 + (z-3)^2 = 3^2$$ and $$(x-3/4)^2 + (y-3/4)^2 + (z-3/4)^2 = (3/4)^2$$ Suppose that the center is $(r, -r, r)$. Then we must have $$\left|\frac{2r-r+2r-6}{3}\right| = r$$ and hence $(r-2)^2 = r^2$ and $r=1$. Thus the sphere $$(x-1)^2 + (y+1)^2 + (z-1)^2 = 1$$ also satisfies the conditions. Similarly we can discuss the other cases.
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I get $7$ solutions in all, considering all octants.
Here is a picture
(it doesn't want to load here, being over 2.7MB).
The centre of your sphere is a point $(\pm r, \pm r, \pm r)$, so that it is a distance $r > 0$ from each of the coordinate planes $x=0$, $y=0$, $z=0$, and is also at distance $r$ from the plane $2x + y + 2z = 6$.
Robert Israel
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