Question: show that the points $(x,y,z)$ which satisfy $x^2 + y^2 + z^2 = 4y-2z$ are a sphere by rewriting this equation in the standard form for a sphere.
I know that the standard form for a sphere is $$(x-a)^2 + (y-b)^2 + (z-c)^2 = R^2.$$
I am not exactly sure how to solve this, I was thinking that the $4y-2z$ are the $b$ and $c$ for the standard form but then would it combine to be $-3y$ and $-z$ or am I completely wrong?
$\endgroup$4 Answers
$\begingroup$You're going to have to complete the square. For example: $$ z^2+2z=0\iff z^2+2z+1=1\iff (z+1)^2=1. $$ Can you do the same for $y$?
Edit: when you do it for $y$, you should get $$ y^2-4y=0\iff y^2-4y+4=4\iff (y-2)^2=4. $$ That'll get you the formula for a circle that you want.
$\endgroup$ 4 $\begingroup$Write $x^2 + y^2 + z^2 = 4y-2z$ as $x^2 + y^2-4y + z^2+2z=0$. Hence,$$x^2 + y^2-4y + z^2+2z$$$$=x^2+y^2-4y+(4-4)+z^2+2x+(1-1)$$ $$=x^2+(y^2-4y+4)-4+(z^2+2x+1)-1$$ $$=x^2+(y-2)^2+(z+1)^2-5$$ so that $x^2+(y-2)^2+(z+1)^2=\sqrt{5}^2$. This is the equation of the sphere of radius $\sqrt{5}$ centered at ${(0,2,-1)}$.
$\endgroup$ $\begingroup$The first step is to group all the terms with the same variable together.
From $x^2 + y^2 + z^2 = 4y-2z$ we get $x^2 + y^2-4y + z^2+2z = 0$.
The next step, as Ian Coley wrote, is to complete the square for each variable. The general formula is $v^2+av =v^2+av + (a/2)^2 - (a/2)^2 =(v+a/2)^2 - a^2/4 $.
In your case, there are two variable for which this needs to be done: $y$ and $z$.
For $y$, since $a=-4$, we get $y^2-4y =(y-2)^2-4 $.
For $z$, since $a=2$, we get $z^2+2z =(z+1)^2-1 $.
Putting these in the equation, it becomes $x^2 + (y-2)^2-4 + (z+1)^2-1 = 0$.
Finally, bringing the constant terms to the right side, we get $x^2 + (y-2)^2 + (z+1)^2 = 4+1=5$.
From this, we see that the sphere has center $(0, 2, -1)$ and radius $\sqrt{5}$.
$\endgroup$ $\begingroup$x^2+y^2+z^2=4y-2z......(1) Now rearranging the terms we get: x^2+y^2-4y+z^2+2z=0 Now we have to make them perfect square x^2+y^2-4y+4-4+z^2+2z+1-1=0 Then We get x^2+(y-2)^2+(z+1)^2-5=0 Or x^2+(y-2)^2+(z+1)^2=5 Therefore it is a sphere having centre (0,2,-1) and radius 5^0.5.
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