Ask Question
12 April, 13:10

Write the equation of the sphere in standard form. X2 + y2 + z2 + 6x - 2y - 6z = 17

+4
Answers (1)
  1. 12 April, 13:58
    0
    We are given equation of sphere x^2 + y^2 + z^2 + 6x - 2y - 6z = 17.

    We know, standard form of a sphere (x-a) ^2 + (y-b) ^2 + (x-c) ^2=r^2.

    Where centered at (a, b, c) (a, b, c) with radius r.

    Let us convert given equation of sphere in standard form by completing the square.

    Writing x, y and z terms seprately.

    (x^2+6x) + (y^2-2y) + (z^2-6z) = 17.

    We have coefficents of x, y and z are 6, - 2 and - 6.

    First we will find half of those coefficents and then square them.

    We get 6/2 = 3, - 2/2 = -1 and - 6/2=-3.

    Now, squaring we get

    (3) ^2 = 9, (-1) ^2 = 1 and (-3) ^2 = 9.

    Adding those numbers inside first, second and third parentheses and adding same numbers on right side.

    (x^2+6x+9) + (y^2-2y+1) + (z^2-6z+9) = 17+9+1+9.

    Now, finding perfect square of each, we get

    (x+3) ^2 + (y-1) ^2 + (z-3) ^2 = 36.

    We could write 36 as 6^2.

    So, final answer would be

    (x+3) ^2 + (y-1) ^2 + (z-3) ^2 = 6^2 is the standard form of a sphere equation.
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “Write the equation of the sphere in standard form. X2 + y2 + z2 + 6x - 2y - 6z = 17 ...” in 📙 Mathematics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers