Solveeit Logo
Login

Question

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

Find the equation of a sphere with centre (2,−6,4)\left( {2, - 6,4} \right) and radius 55 units.

Explanation

Solution

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 rr and centre (a,b,c)\left( {a,b,c} \right), is given by (x−a2)+(y−b)2+(z−c)2=r2\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 (2,−6,4)\left( {2, - 6,4} \right) and radius is 55 units.
We know that the general equation of a sphere with radius rr and centre (a,b,c)\left( {a,b,c} \right), is given by (x−a2)+(y−b)2+(z−c)2=r2\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=4a = 2,b = - 6,c = 4 and r=5r = 5.
Therefore the equation of the sphere will be
(x−2)2+(y+6)2+(z−4)2=52{\left( {x - 2} \right)^2} + {\left( {y + 6} \right)^2} + {\left( {z - 4} \right)^2} = {5^2}
Simplifying the above equation, we get:
x2−4x+4+y2+12y+36+z2−8z+16=25{x^2} - 4x + 4 + {y^2} + 12y + 36 + {z^2} - 8z + 16 = 25
⇒x2+y2+z2−4x+12y−8z+31=0\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 (2,−6,4)\left( {2, - 6,4} \right) and radius 55 is x2+y2+z2−4x+12y−8z+31=0{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 (h,k)\left( {h,k} \right) and radius zz is given by (x−h)2+(y−k)2=z2{\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 rr and centre (a,b,c)\left( {a,b,c} \right), is given by (x−a2)+(y−b)2+(z−c)2=r2\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 x2+y2+z2=r2{x^2} + {y^2} + {z^2} = {r^2}, has its centre at the origin and radius rr.