Question

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

Answer 1

Hint: 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 \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}={{2}^{2}}\].