Question

Find the equation of a sphere with centre \[\left( {2, - 6,4} \right)\] and radius \[5\] units.

Answer 1

Hint: In the given question, the centre of the sphere and also the radius of the sphere is given to us. The standard equation of a sphere with radius \[r\] and centre \[\left( {a,b,c} \right)\], is given by \[\left( {x - {a^2}} \right) + {\left( {y - b} \right)^2} + {\left( {z - c} \right)^2} = {r^2}\]. Try and use this relation to find the equation of the required sphere.

Complete step by step solution:
In the given question we have to find the equation of a sphere whose centre is given as \[\left( {2, - 6,4} \right)\] and radius is \[5\] units.
We know that the general equation of a sphere with radius \[r\] and centre \[\left( {a,b,c} \right)\], is given by \[\left( {x - {a^2}} \right) + {\left( {y - b} \right)^2} + {\left( {z - c} \right)^2} = {r^2}\]. Now comparing with the data of our question, we observe that \[a = 2,b = - 6,c = 4\] and \[r = 5\].
Therefore the equation of the sphere will be
 \[{\left( {x - 2} \right)^2} + {\left( {y + 6} \right)^2} + {\left( {z - 4} \right)^2} = {5^2}\]
Simplifying the above equation, we get:
\[{x^2} - 4x + 4 + {y^2} + 12y + 36 + {z^2} - 8z + 16 = 25\]
\[ \Rightarrow {x^2} + {y^2} + {z^2} - 4x + 12y - 8z + 31 = 0\]
This is the required equation of the sphere.
Thus the equation of the sphere with centre \[\left( {2, - 6,4} \right)\] and radius \[5\] is \[{x^2} + {y^2} + {z^2} - 4x + 12y - 8z + 31 = 0\].

Note: Observe that the equation of the sphere is similar to that of the circle. The equation of a circle of centre \[\left( {h,k} \right)\] and radius \[z\] is given by \[{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {z^2}\]. Since the sphere is a three dimensional object and the circle is a two dimensional object, so there is an extra term in the equation of a sphere to account for the third dimension. Thus the equation of the sphere with radius \[r\] and centre \[\left( {a,b,c} \right)\], is given by \[\left( {x - {a^2}} \right) + {\left( {y - b} \right)^2} + {\left( {z - c} \right)^2} = {r^2}\]. Also note that a sphere having the equation \[{x^2} + {y^2} + {z^2} = {r^2}\], has its centre at the origin and radius \[r\].