Question

How do I find the equation of the sphere of radius 2 centred at the origin?

Answer 1

In order to find the solution of the given question that is to find the equation of the sphere of radius 2 centred at the origin use the Pythagoras theorem which states that in a right-angled triangle the square of the hypotenuse is equal to the sum of the square of two other sides.

Complete step by step solution:
According to the question, given radius of the sphere is as follows:
$2$
Apply the Pythagoras theorem which states that in a right-angled triangle the square of the hypotenuse is equal to the sum of the square of two other sides that is
$Hypotenus{{e}^{2}}={{a}^{2}}+{{b}^{2}}$ where $a$ and $b$ are the length of the two other sides in a right-angled triangle.
We are given that we have to find the equation of the sphere which is centred at origin. Which implies the points $\left( 0,0,0 \right),\left( x,0,0 \right)$ and $\left( x,y,0 \right)$ form the vertices of a right-angled triangle with sides of length $x,y$ and $\sqrt{{{x}^{2}}+{{y}^{2}}}$.
Then the points $\left( 0,0,0 \right)$, $\left( x,y,0 \right)$ and $\left( x,y,z \right)$ form the vertices of a right-angled triangle with sides of length $\sqrt{{{x}^{2}}+{{y}^{2}}}$, $z$ and
$\sqrt{{{\left( \sqrt{{{x}^{2}}+{{y}^{2}}} \right)}^{2}}+{{z}^{2}}}=\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}$
So, we can write the equation of a sphere as:
$\Rightarrow \sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}=2$
After simplifying it further we will have:
$\Rightarrow {{x}^{2}}+{{y}^{2}}+{{z}^{2}}={{2}^{2}}$

Therefore, the required equation of the sphere of radius 2 centred at the origin is ${{x}^{2}}+{{y}^{2}}+{{z}^{2}}={{2}^{2}}$.

Note: There’s an alternative way to find the solution of the given question:
First, we are given the centre of the sphere: $\left( 0,0,0 \right)$ and the radius is equal to $2$. Let's substitute this point into our sphere equation.
$\Rightarrow {{\left( x-h \right)}^{2}}+{{\left( y-k \right)}^{2}}+{{\left( z-l \right)}^{2}}=\text{radiu}{{\text{s}}^{2}}$
Here $\left( h,k,l \right)$ is the centre point of the sphere.
$\Rightarrow {{\left( x-0 \right)}^{2}}+{{\left( y-0 \right)}^{2}}+{{\left( z-0 \right)}^{2}}={{\text{2}}^{2}}$
$\Rightarrow {{x}^{2}}+{{y}^{2}}+{{z}^{2}}={{2}^{2}}$
Therefore, the required equation of the sphere of radius 2 centred at the origin is ${{x}^{2}}+{{y}^{2}}+{{z}^{2}}={{2}^{2}}$.