1. ## calc 3 problem...

complete the square to write the equation of the sphere in standard form. find the center and the radius...

x^2+y^2+z^2-4x-6y+4=0

i dont know how to complete the square like this...can anyone help????

2. Originally Posted by chris25
complete the square to write the equation of the sphere in standard form. find the center and the radius...

x^2+y^2+z^2-4x-6y+4=0

i dont know how to complete the square like this...can anyone help????
is that just a 4? should it be 4z?

just group like terms and complete the square. so complete the squares for x^2 - 4x and y^2 - 6x separately. as for the z's, i am waiting on your response.

this is not a calculus 3 problem

3. Equation of a sphere is given as: $(x-a)^2 + (y-b)^2 + (z-c)^2 = r^2$
where (a,b,c) is the centre of the sphere.

${\color{red}x^{2}} + {\color{blue}y^2} + {\color{magenta}z^2} - {\color{red}4x} - {\color{blue}6y} + 4 = 0$
${\color{red}x^{2} - 4x} + {\color{blue}y^{2} - 6y} + {\color{magenta}z^{2}} = -4$

Now complete the square in respect to each variable:
$\left[x^{2} - 4x + {\color{red}\left(\frac{4}{2}\right)^{2}}\right] + \left[y^{2} - 6y + {\color{blue}\left(\frac{6}{2}\right)^{2}}\right] + z^{2} = -4 + {\color{red}\left(\frac{4}{2}\right)^{2}} + {\color{blue}\left(\frac{6}{2}\right)^{2}}$

etc. etc.

4. its just a 4

5. What does that mean?

6. Originally Posted by o_O
What does that mean?
it is in response to the question i asked

7. Originally Posted by chris25
complete the square to write the equation of the sphere in standard form. find the center and the radius...

x^2+y^2+z^2-4x-6y+4=0

i dont know how to complete the square like this...can anyone help????
We need to group together all x,y, and z's

$(x^2-4x \\\ \\\ )+(y^2-6y \\\ \\\ ) + z^2=-4$

Now we are going to add the square of half the linear coeffient to each side of the equation

$(x^2-4x+4)+(y^2-6y+9)+z^2=-4+4+9$

now we can factor to get

$(x-2)^2+(y-3)^2+z^2=3^2$

So we have a shpere with radius 3 centered at (2,3,0)

Good luck.