**$x^2+y^2+z^2+8x-6y+2z+17=0$**

The main objective of this question is to prove that the **given equation** is for a** sphere** and also to find the **center** and **radius** for a given sphere equation.

This question uses the concept of the **sphere**. A sphere is a **round,** **three-dimensional** object like a ball or moon where each **point** on its surface has an **equal distance** from its center. One of the **properties** of the sphere is that it is perfectly **symmetrical** and it’s not polyhedron. The other property of the **sphere** is its **mean curvature, and circumference and width** are **constant.**

## Expert Answer

The** given** equation is:

\[=x^2+y^2+z^2+8x-6y+2z+17=0\]

We have to prove, that it is a **sphere equation** and finds the **center and radius **of the given sphere equation.

Imagine a sphere with its **centre** $C(h,j,l)$ and its** radius** $r$.

We have **formula **for** sphere **as:

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

where $(h,k,l)$ is the **sphere center** and its radius is represented by $r$.

**Rearranging** the given equation results in:

\[(x^2 +8x +4^2 -4^2)+(y^2-6y+3^2-3^2)+(z^2+2z-1^2-1^2)+17=0\]

\[(x^2 +8x +4^2 )+(y^2-6y+3^2)+(z^2+2z-1^{20})+(-16-9-1)+17=0\]

\[(x^2 +8x +4^2 )+(y^2-6y+3^2)+(z^2+2z-1^{20})+(-16-10)+17=0\]

\[(x^2 +8x +4^2 )+(y^2-6y+3^2)+(z^2+2z-1^{20})+(-26)+17=0\]

\[(x^2 +8x +4^2 )+(y^2-6y+3^2)+(z^2+2z-1^{20})-26+17=0\]

**Moving** $-26$ to the **right side** results in:

\[(x^2 +8x +4^2 )+(y^2-6y+3^2)+(z^2+2z-1^{20})+17=26\]

By **shifting** $17$ to the right side **results** in:

\[(x^2 +8x +4^2 )+(y^2-6y+3^2)+(z^2+2z-1^{20})+=26-17\]

**Subtracting** the **right side** term results in:

\[(x^2 +8x +4^2 )+(y^2-6y+3^2)+(z^2+2z-1^{20})=9\]

\[[x-(-4)]^2 +(y-3)^2 +[z-(-1)]^2=(3)^2(2)\]

Now **comparing** the two-equations, we get:

$h$=-4.

$k$=3.

$l$=-1.

$r$=3.

Therefore, the **sphere center** is $(-4,3,1)$ and its **radius** is $3$.

## Numeric Answer

For the **given sphere equation, **it is proved that it’s of the sphere and the **center** is $(-4,3,1)$, with a **radius** of $3$.

## Example

Show that the given two equations are for the sphere and also find the center and radius for these two-sphere equations.

\[2x^2+2y^2+2z^2=8x-24z+1\]

\[x^2+y^2+z^2-2x-4y+8z=15\]

Imagine a sphere with its **centre** $C(h,j,l)$ and its **radius** $r$. It is represented by **formula** as:

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

where $(h,k,l)$ is the** sphere center** and its **radius** is represented by $r$.

The **given** sphere equation is:

\[2x^2+2y^2+2z^2=8x-24z+1\]

**Dividing** the given equation by $2$ results in:

\[x^2-4x+y^2+z^2+12z=\frac{1}{2}\]

For a **complete square**, we have to add 40 to both sides.

\[x^2-4x+4+y^2+z^2+12z+36=\frac{1}{2} + 40\]

**Adding** 40 to **both sides** result in:

\[x^2-4x+y^2+z^2+12z+40=\frac{81}{2}\]

Make a **square term **so that we can **compare** it with the equation of a **sphere**.

\[(x-2)^2 +y^2 +(z+6)^2=\frac{81}{2}\]

Now for the $2^{nd}$, given equation, we have to **prove** its **sphere** equation and also to find the **center and radius **of this equation.

\[(x^2+2x)+(y^2+4y)+(z^2+8z)=15\]

By **simplifying** the given equation,we get:

\[(x-1)^2 +(y-2)^2 +(z-(-4)^2)=6^2\]

Now, this **equation** is in the form of a **standard sphere** equation. By **comparing** this equation with the standard sphere equation **results** in:

$center=(1,2,-4)$

$radius=6$

**Hence,** it is **proved** that the **given equation** is for sphere with **center** $(2,0,-6)$ and **radius** $\frac{9}{\sqrt{2}}$ and for the $2^{nd}$ equation $x^2+y^2+z^2-2x-4y+8z=15$ is also for **sphere** and its **center** is $(1,2,-4)$ and **radius** is $6$.