Hello everyone, today I will show you how to write the equation of a sphere in space!

Similar to **how to write the equation of a circle on the plane**I will also guide you to the two most common cases (that is, knowing the center and radius, knowing the four passing points).

Before starting to learn, if possible, prepare your CASIO fx-580VN X computer first (link below), or install an emulator on your computer. We will need it to quickly solve a system of quadratic equations of four unknowns!

**#first. Write the equation of the sphere when the center and radius are known**

It has been shown that an equation of a sphere is completely definite when its center and radius are known.

The equation of the sphere with center $I=(x_0; y_0; z_0)$ and radius R would be $(x-x_0)^2+(y-y_0)^2+(z-z_0)^2=R^2 $

**Attention:**The equation of the sphere passing through the origin $O=(0; 0; 0)$ and the radius

**CHEAP**would be $x^2+y^2+z^2=R^2$

**Example 1.** Write an equation of a sphere with center $I=(2; 3; 5)$ and radius $R=7$

**The answer:**

The required sphere equation has the form $(x-x_0)^2+(y-y_0)^2+(z-z_0)^2=R^2$

Since this equation has center $I=(2; 3; 5)$ and radius $R=7$, we get $(x-2)^2+(y-3)^2+(z-5) ^2=49$

So => the required sphere equation is $(x-2)^2+(y-3)^2+(z-5)^2=49$

**Example 2.** Write the equation of the sphere passing through the two points $A=(2; 3; 5)$ and $B=(7; 11; 13)$ and take the line segment AB as the diameter.

**The process of finding the answer:**

Since the required sphere equation passes through two points A and B and takes the line segment AB as the diameter, we have…

- The coordinates of the midpoints of two points A and B are the centers of the sphere.
- Half the length of line segment AB is the radius.

**The answer:**

The coordinates of the center of the sphere to find are $\left(\frac{2+7}{2}; \frac{3+11}{2}; \frac{5+13}{2}\right)$ or $\left(\frac{9}{2}; 7; 9\right)$

The radius of the sphere to be found is $\frac{\sqrt{(7-2)^2+(11-3)^2+(13-5)^2}}{2}=\frac{3\sqrt {17}}{2}$

So => the required sphere equation is $\left(x-\frac{9}{2}\right)^2+(y-7)^2+(z-9)^2=\frac{153 }{4}$

**Comment:**

- Suppose I change my hypothesis a bit: “Write the equation of the sphere passing through two points $A=(2; 3; 5)$ and $B=(7; 11; 13)$ and take the line segment AB as
**radius**” then we will be able to write two spheres satisfying the requirements of the problem. - Nature of
**Example 2**is also to write the equation of the sphere when the center and radius are known.

**#2. When you know the four points that pass through**

The sphere equation $(x-x_0)^2+(y-y_0)^2+(z-z_0)^2=R^2$ can also be written as an expansion $x^2+y^2 +z^2-2x_0x-2y_0y-2z_0z+d=0$ with $d=x_0^2+y_0^2+z_0^2-R^2$

When asked to write the equation of a sphere passing through four points we will use the equation in expanded form

Suppose the required sphere equation passes through four points $A=(x_a; y_a; z_a), B=(x_b; y_b; z_b), C=(x_c; y_c; z_c), D=(x_d; y_d; z_d)$

**Step 1.** Substituting the coordinates of the four points A, B, C, D in the expansion equation, respectively, we will get a system of four quadratic equations with four unknowns:

$\left\{\begin{array}{ll}x_a^2+y_a^2+z_a^2-2x_0x_a-2y_0y_a-2z_0z_a+d&=0\\x_b^2+y_b^2+z_b^2-2x_0x_b- 2y_0y_b-2z_0z_b+d&=0\\x_c^2+y_c^2+z_c^2-2x_0x_c-2y_0y_c-2z_0z_c+d&=0\\x_d^2+y_d^2+z_d^2-2x_0x_d-2y_0z_d-2+z d&=0\end{array}\right.$

**Step 2.** Use method substitution methodthe addition method, … to solve this system we will find the solution $x_0, y_0, z_0, d$

**Step 3.** Substituting $x_0, y_0, z_0, d$ into the equation $x^2+y^2+z^2-2x_0x-2y_0y-2z_0z+d=0$ we get a sphere equation

**Attention:**

- Phrases
**four points on the sphere**,**four points on the sphere**,**the sphere passes through four points**is the same meaning - There are infinitely many equations of the sphere passing through the three points

**Example 3.** Write the equation of the sphere passing through the four points $A(2; 4; -1), B(1; 4; -1), C(2; 3; 4), D(2; 2; -1)$

**The answer:**

The required sphere equation has the form $x^2+y^2+z^2-2x_0x-2y_0y-2z_0z+d=0$ and passes through four points $A(2; 4; -1), B(1 ; 4; -1), C(2; 3; 4), D(2; 2; -1)$ so we get the system of equations:

$\left\{\begin{array}{ll}2^2+4^2+(-1)^2-2x_02-2y_04-2z_0(-1)+d&=0\\1^2+4^2 +(-1)^2-2x_01-2y_04-2z_0(-1)+d&=0\\2^2+3^2+4^2-2x_02-2y_03-2z_04+d&=0\\2^2+ 2^2+(-1)^2-2x_02-2y_02-2z_0(-1)+d&=0\end{array}\right. \Leftrightarrow \left\{\begin{array}{ll}-4x_0-8y_0+2z_0+d&=-21\\-2x_0-8y_0+2z_0+d&=-18\\-4x_0-6y_0-8z_0+d&=- 29\\-4x_0-4y_0+2z_0+d&=-9\end{array}\right. \Leftrightarrow \left\{\begin{array}{ll}x_0&=\frac{3}{2}\\y_0&=3\\z_0&=\frac{7}{5}\\d&=\frac{31} {5}\end{array}\right.$

So => the equation for the demand surface is $x^2+y^2+z^2-3x-6y-\frac{14}{5}z+\frac{31}{5}=0$

**#3. Quick trick to write the equation of a sphere passing through four points using a CASIO . calculator**

The foundation of this trick is the equation $x^2+y^2+z^2-2x_0x-2y_0y-2z_0z+d=0$ and the calculation method **Equation / Func**

**Attention:**

Only calculator CASIO fx-580VN X The new feature has the ability to solve a system of quadratic quadratic equations, if you use the calculator CASIO fx-570VN Plus, CASIO fx-570ES Plus, … then you cannot use this trick.

**Example 4.** The equation of the sphere passing through the four points $A(2; 4; -1), B(1; 4; -1), C(2; 3; 4), D(2; 2; -1)$ is

- $x^2+y^2+z^2-3x-6y-\frac{14}{5}z+\frac{31}{5}=0$
- $x^2+y^2+z^2+3x-6y-\frac{14}{5}z+\frac{31}{5}=0$
- $x^2+y^2+z^2-3x+6y-\frac{14}{5}z+\frac{31}{5}=0$
- $x^2+y^2+z^2-3x-6y+\frac{14}{5}z+\frac{31}{5}=0$

**Implementation steps:**

**Step 1.** Select calculation method `Equation / Func`

**Step 2.** Choose `Simul Equation`

**Step 3.** Press the number key `4`

on the keyboard

**Step 4.** Enter the coefficients of the system of equations…

- Opposite of the first equation the values of the coefficients in order will be
**twice the first point**=>**twice the median of the first point**=>**twice the altitude of the first point**=>**-first**=>**sum of the squares of the degrees, mid, and altitudes of the first point** - The second, third, and fourth equations are similar.

**Step 5.** Press `=`

=> press `=`

=> press `=`

=> press `=`

So the required sphere equation is $x^2+y^2+z^2-3x-6y-\frac{14}{5}z+\frac{31}{5}=0$

The answer is the plan `A`

**#4. Epilogue**

The equation of a circle, the equation of a sphere, is a set of points equidistant from a given point with a given fixed distance.

These two equations differ only in that, the equation of the circle lies in the plane, and the equation of the sphere lies in space. That’s all !

It can be said that the equation of the sphere is an extension of the equation of the circle. If you keep expanding you will get an equation called the hypersphere equation..

Yes, those are a couple of words to expand knowledge instead of the conclusion that I want to send to you. Goodbye and see you in the next posts ha!

**CTV: Nhut Nguyen** – Blogchiasekienthuc.com

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