Equation Of A Sphere

Question

Find the radius of the sphere whose equation is x^2 + y^2 + z^2 - 2x + 4y - 6z + 5 = 0 in three-dimensional space.

RankGuru · Accepted Answer

The method is completing the square to convert the general sphere x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d = 0 into center-radius form, where the center is (-u, -v, -w) and the radius is \sqrt{u^2 + v^2 + w^2 - d}. This conversion is a standard JEE Advanced sphere technique. Matching coefficients: 2u = -2 so u = -1; 2v = 4 so v = 2; 2w = -6 so w = -3; and d = 5. The radius squared is u^2 + v^2 + w^2 - d = 1 + 4 + 9 - 5 = 9, hence the radius is \sqrt{9} = 3, with center (1, -2, 3). Option \sqrt{14} forgets to subtract d. Option 9 reports the radius squared instead of the radius. Option \sqrt{5} misuses the constant term as if it were the radius squared. This applies the completing-the-square sphere theorem. Plausibility check: the radius is positive and real because u^2 + v^2 + w^2 - d = 9 > 0, confirming a genuine non-degenerate sphere rather than a point or imaginary locus.

Equation Of A Sphere

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