Question:
The radius of the sphere whose centre is \((4,4,-2)\) and which passes through origin is
The radius of the sphere whose centre is \((4,4,-2)\) and which passes through origin is
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If a sphere passes through origin, its radius is the distance between its centre and origin.
Updated On: Jul 3, 2026
- \(\sqrt2\)
- \(3\)
- \(2\sqrt2\)
- \(6\)
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The Correct Option is D
Solution and Explanation
Concept:
The radius of a sphere is the distance from its centre to any point on the sphere. Here, the sphere passes through the origin. So the radius is the distance between the centre \((4,4,-2)\) and the origin \((0,0,0)\).
Step 1: Use distance formula. Distance between \[ (x_1,y_1,z_1) \] and \[ (x_2,y_2,z_2) \] is \[ d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2} \]
Step 2: Substitute the points. Centre: \[ (4,4,-2) \] Origin: \[ (0,0,0) \] So, \[ r=\sqrt{(4-0)^2+(4-0)^2+(-2-0)^2} \] \[ r=\sqrt{16+16+4} \] \[ r=\sqrt{36} \] \[ r=6 \]
Step 3: Final answer. \[ \boxed{6} \]
The radius of a sphere is the distance from its centre to any point on the sphere. Here, the sphere passes through the origin. So the radius is the distance between the centre \((4,4,-2)\) and the origin \((0,0,0)\).
Step 1: Use distance formula. Distance between \[ (x_1,y_1,z_1) \] and \[ (x_2,y_2,z_2) \] is \[ d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2} \]
Step 2: Substitute the points. Centre: \[ (4,4,-2) \] Origin: \[ (0,0,0) \] So, \[ r=\sqrt{(4-0)^2+(4-0)^2+(-2-0)^2} \] \[ r=\sqrt{16+16+4} \] \[ r=\sqrt{36} \] \[ r=6 \]
Step 3: Final answer. \[ \boxed{6} \]
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