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What is the equation to the sphere whose centre is at (-2, 3, 4) and radius is 6 units?
1. x2 + y2 + z2 + 4x – 6y – 8z = 7
2. x2 + y2 + z2 + 6x – 4y – 8z = 7
3. x2 + y2 + z2 + 4x – 6y – 8z = 4
4. x2 + y2 + z2 + 4x + 6y + 8z = 4
5. None of these

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Correct Answer - Option 1 : x2 + y2 + z2 + 4x – 6y – 8z = 7

Concept:

  • General equation of a sphere cantered at the point (x0, y0, z0) with radius R is given by
  • (x – x0)2 + (y – y0)2 + (z – z0)2 = R2


Calculation:

Given: Centre is at (-2, 3, 4) and radius is 6

⇒ (x0, y0, z0) = (-2, 3, 4) and R = 6

We know that general equation of sphere is (x – x0)2 + (y – y0)2 + (z – z0)2 = R2

⇒ (x – (-2)) 2 + (y – 3)2 + (z – 4)2 = 62

⇒ (x + 2) 2 + (y – 3)2 + (z – 4)2 = 62

⇒ x2 + 4x + 4 + y2 – 6y + 9 + z2 – 8z + 16 = 36

⇒ x2 + y2 + z2 + 4x – 6y – 8z + 29 = 36

⇒ x2 + y2 + z2 + 4x – 6y – 8z = 7

∴ Option 1 is correct.

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