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I am given the equation $3x^2 - 9x + 3y^2 + 7y + 3z^2 - 4z + 2 = 0$ I started by dividing by three to get a leading coefficient of 1 which gives $x^2 - 3x + y^2 + \frac{7}{3}y + z^2 - \frac{4}{3}z = -\frac{2}{3}$ from here I try to complete the square which gives $ (x - \frac{3}{2})^2 + (y + 7/6)^2 + (z - \frac{4}{6})^2 = -\frac{2}{3} + \frac{9}{4} + \frac{49}{36} + \frac{16}{36}$ reducing the right side I get $ (x - \frac{3}{2})^2 + (y + 7/6)^2 + (z - \frac{4}{6})^2 = \frac{122}{36}$ Is this the correct equation and how do i find/use the radius and center to sketch the sphere.

Doug Ray
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  • That is the right approach (and the arithmetic looks right). Once you have it in this form $(x-x_0)^2 + (y-y_0)^2 + (z-z_0)^2 = R^2$ the center is $(x_0,y_0,z_0)$ and the radius is $R$ – Doug M Sep 06 '17 at 18:13
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    i have got $$\left(x-\frac{3}{2}\right)^2+\left(y+\frac{7}{6}\right)^2+\left(z-\frac{2}{3}\right)^2=\frac{122}{108}$$ – Dr. Sonnhard Graubner Sep 06 '17 at 18:14
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    $$\left(x-\frac{3}{2}\right)^2+\left(y+\frac{7}{6}\right)^2+\left(z-\frac{2}{3}\right)^2=\left(\frac{1}{3}\sqrt{\frac{61}{2}}\right)^2$$ $radius=\dfrac{1}{3}\sqrt{\dfrac{61}{2}}\approx 1.84$ – Raffaele Sep 06 '17 at 18:25

2 Answers2

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It seems like your approach is correct. If you know the center of the sphere, and the radius, clearly you know what every point on the sphere should look like. In terms of actually sketching, it will help to first draw the equator, where if $c=(a,b,z_0)$ is the center, draw the circle $(x-a)^2+(y-b)^2=r^2-z_0^2$.

On the other hand, completing the square is a good idea. I usually just split this up into a few different equations heuristically:

$3x^2-9x$ needs to be completed. But $9=3*2*b$ so $b=3/2$. Hence, we ad $9/4$ to both sides of the equation etc.

Continuing in this way, we obtain the equation $$(x-3/2)^2+(y+7/6)^2+(z-2/3)^2=-2/3+9/4+49/36+4/9=61/18$$

More generally, $(x-a)^2+(y-b)^2+(z-c)^2=r^2$ will be a sphere of radius $r$ with center $(a,b,c)$.

Andres Mejia
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Here’s a different, but easy way to do this:

Given the general equation $q(x,y,z)=0$ of a quadric that has a center, you can find the center by computing where the gradient vanishes. For this sphere, this means solving $\nabla q=(6x-9,6y+7,6z-4)=0$, from which the center is $c=\left(\frac32,-\frac76,\frac23\right)$.

To get the radius, plug the coordinates of the center into the left-hand side of the sphere’s equation and divide by the common coefficient of the quadratic terms. The result is the negative of the radius squared. Plugging $c$ into the equation at hand, we get $$3\left(\frac32\right)^2-9\left(\frac32\right)+3\left(-\frac76\right)^2+7\left(-\frac76\right)+3\left(\frac23\right)^2-4\left(\frac23\right)+2 = -{61\over6}.$$ Dividing by $3$ and taking the square root gives $\frac13\sqrt{{61\over2}}$ for the radius.

See this question for some explanations of why taking partial derivatives to find the center works. Those explanations are for two dimensions, but generalize. As for the method for finding the radius, consider what happens when we translate the general equation $$q(x,y,z)=A(x^2+y^2+z^2)+Bx+Cy+Dz+E=0$$ of a sphere by $(x_0,y_0,z_0)$, i.e., make the replacements $x\mapsto x'-x_0$, $y\mapsto y'-y_0$, $z\mapsto z'-z_0$. The new constant term of the transformed equation is $$A(x_0^2+y_0^2+z_0^2)-Bx_0-Cy_0-Dz_0+E=q(-x_0,-y_0,-z_0).$$ If we translate so that the sphere’s center is at the origin, the linear terms of the equation vanish, so we’re left with $$A(x'^2+y'^2+z'^2)+q(x_c,y_c,z_c)=0$$ or $$x'^2+y'^2+z'^2=-\frac1Aq(x_c,y_c,z_c).$$ This holds more generally: if you translate the equation of a quadric $q(\mathbf x)=0$ by $\mathbf v$, the constant term of the transformed equation is $q(-\mathbf v)$. Knowing this can save having to do some tedious calculations.

amd
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