Question

Find the equation of a sphere, whose end points of a diameter are (0,0,0) and (1,2,3)

• A. $x^2+y^2+z^2-x-2y-3z=0$
• B. $x^2+y^2+z^2-2x-y-3z=0$
• C. $x^2+y^2+z^2-3x-2y-z=0$
• D. $x^2+y^2+z^2-x-2y-z=0$

Answer 1

The correct option is A

$x^2+y^2+z^2-x-2y-3z=0$

Equation of a sphere whose extremities of diameter are
$A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$ is

$(x–x_1)(x–x_2)+(y–y_1)(y–y_2)+(z–z_1)(z–z_2)=0.$

We can substitute the values and find the equation.

$\Rightarrow (x-0)(x-1)+(y-0)(y-2)+(z-0)(z-3)=0\Rightarrow x^2+y^2+z^2-x-2y-3z=0$

This is the equation of the sphere

If we don’t know the above formula, then the following method can be used to find the equation of sphere

1. Find the centre of the sphere : Centre will be the midpoint of the endpoints of diameter

2. Find the radius : Distance between centre and any of the points given will be equal to radius

3. Find the equation using centre and radius : $(x-a{)}^2+(y-b{)}^2+(z-c{)}^2=(r{)}^2$ , ,

where (a,b,c) is the centre and r is the radius