### Shortcut Attempt.

Sep. 9th, 2010 06:02 pm**netquiddler**

Yesterday, in calculus III class, we learned that the formula of a sphere is given by:

[;(x-h)^2+(y-k)^2+(z-l)^2=r^2;]

where the center is at [;(h,k,l);] and the radius is [;r;]. Now, my professor has decided to give us problems like this:

[;x^2+y^2+z^2-4x+6y-2z+6=0;]

asking whether this is a formula for a sphere, and if so, what is its center and radius. This problem would be solved like this.

[;x^2+y^2+z^2+4x-6y+2z+6=0;]

[;x^2+4x+y^2-6y+z^2+2z=-6;]

[;x^2+4x+4+y^2-6y+9+z^2+2z+1=-6+4+9+1;]

[;(x+2)^2+(y-3)^2+(z-1)^2=8;]

This shows a sphere with a radius of [;2\sqrt{2};] and a center of [;(-2,3,-1);]. (We would not have a sphere if we ended up with a nonpositive radius.)

Now, rather than having to trudge through that work every time we do this problem, I think I'm going to be clever.

Suppose we have a problem in this form:

[;x^2+y^2+z^2+Ax+By+Cz+D=0;]

We can do the same thing to this problem.

[;x^2+y^2+z^2+Ax+By+Cz+D=0;]

[;x^2+Ax+y^2+By+z^2+Cz=-D;]

[;x^2+Ax+\left(\frac{A}{2}\right)^2+y^2+By+\left(\frac{B}{2}\right)^2+z^2+Cz+\left(\frac{C}{2}\right)^2=\left(\frac{A}{2}\right)^2+\left(\frac{B}{2}\right)^2+\left(\frac{C}{2}\right)^2-D;]

[;\left(x+\frac{A}{2}\right)^2+\left(y+\frac{B}{2}\right)^2+\left(z+\frac{C}{2}\right)^2=\frac{A^2+B^2+C^2}{4}-D;]

This leaves us a sphere if and only if [;\frac{A^2+B^2+C^2}{4}>D;]. Now if [;D;] is negative, then this will always be true. If [;D;] is zero, then [;A;], [;B;], and [;C;] would all have to be zero for this to be false.

Any sphere, then, has a center of [;\left(-\frac{A}{2},-\frac{B}{2},-\frac{C}{2}\right);] and a radius of [;\frac{\sqrt{A^2+B^2+C^2-4D}}{2};].

I think this is a valid shortcut. What say the rest of you?

[To see formulas: http://thewe.net/tex]

[;(x-h)^2+(y-k)^2+(z-l)^2=r^2;]

where the center is at [;(h,k,l);] and the radius is [;r;]. Now, my professor has decided to give us problems like this:

[;x^2+y^2+z^2-4x+6y-2z+6=0;]

asking whether this is a formula for a sphere, and if so, what is its center and radius. This problem would be solved like this.

[;x^2+y^2+z^2+4x-6y+2z+6=0;]

[;x^2+4x+y^2-6y+z^2+2z=-6;]

[;x^2+4x+4+y^2-6y+9+z^2+2z+1=-6+4+9+1;]

[;(x+2)^2+(y-3)^2+(z-1)^2=8;]

This shows a sphere with a radius of [;2\sqrt{2};] and a center of [;(-2,3,-1);]. (We would not have a sphere if we ended up with a nonpositive radius.)

Now, rather than having to trudge through that work every time we do this problem, I think I'm going to be clever.

Suppose we have a problem in this form:

[;x^2+y^2+z^2+Ax+By+Cz+D=0;]

We can do the same thing to this problem.

[;x^2+y^2+z^2+Ax+By+Cz+D=0;]

[;x^2+Ax+y^2+By+z^2+Cz=-D;]

[;x^2+Ax+\left(\frac{A}{2}\right)^2+y^2+

[;\left(x+\frac{A}{2}\right)^2+\left(y+\

This leaves us a sphere if and only if [;\frac{A^2+B^2+C^2}{4}>D;]. Now if [;D;] is negative, then this will always be true. If [;D;] is zero, then [;A;], [;B;], and [;C;] would all have to be zero for this to be false.

Any sphere, then, has a center of [;\left(-\frac{A}{2},-\frac{B}{2},-\

I think this is a valid shortcut. What say the rest of you?

[To see formulas: http://thewe.net/tex]