NIT Puducherry Seat Matrix 2025 - Check Previous Year Matrix Here

Equation of Sphere - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • 8 Questions around this concept.

Solve by difficulty

The radius of the circle in which the sphere $x^2+y^2+z^2+2 x-2 y-4 z-19=0$ is cut by the plane $x+2 y+2 z+7=0$ is

 

 

Find the radius of circular section in which the sphere |r| = 5 is cut by the plane $r.(\widehat{i}+\widehat{j}+\widehat{k}) - 3\sqrt3 = 0$

Concepts Covered - 1

Equation of Sphere

A Sphere is the locus of a point which moves in space in such a way that its distance from a fixed point always remains constant. The fixed point is called the centre of the sphere and the fixed distance is called the radius of the sphere.

Equation of Sphere

Sphere's are the 3D representations of circles. The equation of a sphere is similar to that of a circle but with an extra variable for the extra dimension.

Let C(a, b, c) be the centre of the sphere with radius r. Let P(x, y, z) be any point on the sphere,

Then,

$
\begin{aligned}
C P & =r \\
|C P| & =r
\end{aligned}
$

From the distance formula
$
\begin{array}{cc}
\Rightarrow & C P^2=r^2 \\
\Rightarrow & (x-a)^2+(y-b)^2+(z-c)^2=r^2
\end{array}
$
since, $P(x, y, z)$ is an arbitrary point on the sphere, therefore required equation of the sphere is
$
(x-a)^2+(y-b)^2+(z-c)^2=r^2
$

NOTE:
If the centre of the sphere at the origin the equation of the sphere is $x^2+y^2+z^2=r^2$.
The equation of the sphere can also be written as $x^2+y^2+z^2-2(a x+b y+c z)+d=0$, where $d=a^2+b^2+c^2-r^2$

Study it with Videos

Equation of Sphere

"Stay in the loop. Receive exam news, study resources, and expert advice!"

Get Answer to all your questions

Back to top