Question

Write the equation of the sphere in standard form. x^2 + y^2 + z^2 + 2x − 4y − 6z = 22

Answer 1

The equation $x^{2} +y^{2} +z^{2} +2x-4y-6z=22$  in standard form looks like $(x+1)^{2} +(y-2)^{2} +(z-3)^{2} =(6)^{2}$.

Given equation of sphere be $x^{2} +y^{2} +z^{2} +2x-4y-6z=22$.

We are required to express the given equation in the standard form of the equation of sphere.

Equation is basically relationship between two or more variables that are expressed in equal to form. Equation of two variables look like ax+by=c. It may be linear equation,quadratic equation, cubic equation or many more depending on the powers of variables. The standard form of the equation of sphere looks like  $x^{2} +y^{2} +z^{2} =r^{2}$.

The given equation is $x^{2} +y^{2} +z^{2} +2x-4y-6z=22$.

We have to break 22 which is in right side into various parts according to the left side of the equation.

$x^{2} +y^{2} +z^{2} +2x-4y-6z=-1-4-9+36$

$x^{2} +y^{2} +z^{2} +2x-4y-6z+1+4+9=36$

$x^{2} +1+2x+y^{2}+4-4y+z^{2} +9-6z=36$

$x^{2} +(1)^{2} +2*1*x+y^{2} +(2)^{2} -2*2y+z^{2} +(3)^{2} -2*3z=36$

$(x+1)^{2} +(y-2)^{2} +(z-3)^{2} =(6)^{2}$

Hence the equation $x^{2} +y^{2} +z^{2} +2x-4y-6z=22$  in standard form looks like $(x+1)^{2} +(y-2)^{2} +(z-3)^{2} =(6)^{2}$.

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