x^2 + y^2 + z^2 = 4\sqrt{x^2 + y^2}
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- Recall in spherical coordinates:
- ho^2 = x^2 + y^2 + z^2, \quad
- ho\sin\phi = \sqrt{x^2 + y^2}
- Let $r = \sqrt{x^2 + y^2}$. Then:
- r^2 + z^2 = 4r \Rightarrow r^2 - 4r + z^2 = 0 \Rightarrow (r - 2)^2 + z^2 = 4
- This is a sphere of radius 2 centered at $(2, 0, 0)$ in cylindrical coordinates, which corresponds to a **sphere tangent to the y-z plane** in Cartesian coordinates.
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