NDA Maths · 3D Geometry

The Sphere

A sphere is all points at a fixed distance from a centre; its equation reveals the centre and radius, and a perpendicular distance settles how it meets a plane.

All 3D Geometry notes NDA Maths strategy

Why this matters

Twenty PYQs across 2017–2026 — the chapter's most self-contained subtopic and a reliable scorer. The work is reading centre and radius off the general equation, building spheres from conditions (diameter endpoints, concentric, through a point), and the tangency test against a plane. Six concepts, ending with the HARD locus problems.

Concept 1 of 6

General equation, centre, and radius

Intuition

The general sphere

x^2+y^2+z^2+2ux+2vy+2wz+d=0

hides its centre and radius in the coefficients: the centre is

(-u,-v,-w)

(half the linear coefficients, negated) and the radius comes from

\sqrt{u^2+v^2+w^2-d}

. Note the squared terms must have coefficient 1.

Definition

For $x^2+y^2+z^2+2ux+2vy+2wz+d=0$ : centre $= (-u,-v,-w)$ , radius $= \sqrt{u^2+v^2+w^2-d}$ . Always ensure the coefficients of $x^2, y^2, z^2$ are each 1 (divide through if needed) before reading $u, v, w, d$ .

Centre and radius of a sphere

\text{centre} = (-u, -v, -w), \quad r = \sqrt{u^2 + v^2 + w^2 - d}

Diagram · sphere centre & radius (drag to rotate)

From x² + y² + z² + 2ux + 2vy + 2wz + d = 0: centre C = (−u, −v, −w), radius r = √(u² + v² + w² − d).

Worked example

Find the radius of the sphere

x^2 + y^2 + z^2 - 2x + 4y - 6z - 2 = 0

Match $2u = -2,\ 2v = 4,\ 2w = -6$ : so $u = -1,\ v = 2,\ w = -3$ , and $d = -2$ .
Centre $= (1, -2, 3)$ .
Radius $= \sqrt{(-1)^2 + 2^2 + (-3)^2 - (-2)} = \sqrt{1+4+9+2} = \sqrt{16} = 4$ .

Answer:Radius

= 4

(centre

(1,-2,3)

Practice this conceptself-check · 4 quick reps

Try it yourself

Find the centre and radius of

x^2+y^2+z^2 - 2x - 4y + 2z - 3 = 0

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Centre of $x^2+y^2+z^2-4x-6y-8z-16=0$ ?
2.
Radius formula in terms of u,v,w,d?
3.
Centre of $x^2+y^2+z^2+2x=0$ ?
4.
Must the coefficients of $x^2,y^2,z^2$ be?

From the bank · past-year question

Example 13D GeometryEASY

What is the radius of the sphere

x^2+y^2+z^2-6x+8y-10z+1=0

[Q62 · Apr · 2019]

Drill 6 more on general equation, centre, and radius

Concept 2 of 6

Diameter form of a sphere

Intuition

If you're handed the two ENDS of a diameter, you don't need the centre and radius separately — the diameter form writes the sphere directly. The idea: any point

P

on the sphere sees the diameter at a right angle, so

\overrightarrow{PA}\cdot\overrightarrow{PB}=0

Definition

With diameter endpoints $A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$ , the sphere is $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) + (z-z_1)(z-z_2) = 0$ . The centre is the midpoint of $AB$ and the radius is half of $AB$ .

Diameter form

(x-x_1)(x-x_2) + (y-y_1)(y-y_2) + (z-z_1)(z-z_2) = 0

Worked example

(1,2,3)

and

(3,4,5)

are the endpoints of a diameter of

x^2+y^2+z^2+2ux+2vy+2wz+d=0

. Find

u+v+w

Centre = midpoint of the diameter = $\left(\tfrac{1+3}{2}, \tfrac{2+4}{2}, \tfrac{3+5}{2}\right) = (2, 3, 4)$ .
Centre $= (-u,-v,-w)$ , so $u = -2,\ v = -3,\ w = -4$ .
$u + v + w = -2 - 3 - 4 = -9$ .

Answer:

u + v + w = -9

Practice this conceptself-check · 4 quick reps

Try it yourself

Find the centre and radius of the sphere with diameter endpoints

(2,3,5)

and

(4,9,3)

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Centre of a sphere given its diameter endpoints?
2.
Radius given diameter endpoints?
3.
Diameter ends $(0,0,0)$ , $(2,0,0)$ : centre?
4.
On the diameter form, $\overrightarrow{PA}\cdot\overrightarrow{PB} = ?$

From the bank · past-year question

Example 23D GeometryEASY

(1,-1,2)

and

(2,1,-1)

are the endpoints of a diameter of a sphere

x^2+y^2+z^2+2ux+2vy+2wz-1=0

, then what is

u+v+w

equal to?

[Q52 · Apr · 2024]

Concept 3 of 6

Building a sphere from conditions

Intuition

Most sphere questions hand you conditions instead of the equation: a centre and radius, or 'concentric with this sphere and passing through that point'. Concentric means SAME centre — reuse

u, v, w

and only the constant

d

changes; pin it down with the through-point.

Definition

**Centre $(a,b,c)$ , radius $r$ :** $(x-a)^2 + (y-b)^2 + (z-c)^2 = r^2$ . Concentric spheres share a centre, so they share $u, v, w$ ; only $d$ differs. To make it pass through a given point, substitute that point to solve for $d$ (passing through the origin forces $d = 0$ ).

Worked example

Find the equation of the sphere with centre

(2, -1, 3)

and radius 5.

Centre-radius form: $(x-2)^2 + (y+1)^2 + (z-3)^2 = 25$ .
Expand: $x^2 - 4x + 4 + y^2 + 2y + 1 + z^2 - 6z + 9 = 25$ .
Collect: $x^2 + y^2 + z^2 - 4x + 2y - 6z - 11 = 0$ .

Answer:

x^2 + y^2 + z^2 - 4x + 2y - 6z - 11 = 0

Practice this conceptself-check · 4 quick reps

Try it yourself

Find the sphere concentric with

x^2+y^2+z^2-2x-6y-8z-5=0

and passing through the origin.

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Concentric spheres share their?
2.
Sphere centre origin radius 5?
3.
Passing through the origin forces which constant to?
4.
Centre $(1,0,0)$ , radius 2 — equation?

From the bank · past-year question

Example 33D GeometryEASY

What is the equation of the sphere with centre (-2,3,4) and radius 6?

[Q65 · Apr · 2018]

Drill 3 more on building a sphere from conditions

Concept 4 of 6

Sphere and a plane — tangency and sections

Intuition

Whether a plane misses, touches, or cuts a sphere is decided by ONE number: the perpendicular distance

p

from the centre to the plane, compared with the radius

r

p>r

misses,

p=r

is tangent (touches),

p<r

cuts a circle of radius

\sqrt{r^2-p^2}

Definition

Let $p$ be the perpendicular distance from the sphere's centre to the plane and $r$ the radius:

$p > r$ : no intersection.
$p = r$ : the plane is tangent (touches at one point) — the key NDA condition.
$p < r$ : the plane cuts a circle of radius $\sqrt{r^2 - p^2}$ .

Tangency condition

p = r \quad\text{(perpendicular distance from centre = radius)}

Worked example

Find the diameter of the sphere whose centre is

(1, 2, 2)

and which touches the plane

x + 2y + 2z + 3 = 0

Touches → radius = perpendicular distance from centre to plane.
$p = \dfrac{|1(1) + 2(2) + 2(2) + 3|}{\sqrt{1+4+4}} = \dfrac{|1+4+4+3|}{\sqrt{9}} = \dfrac{12}{3} = 4$ .
So radius $r = 4$ ; diameter $= 8$ .

Answer:Diameter

= 8

Practice this conceptself-check · 4 quick reps

Try it yourself

Does the plane

z = 0

cut, touch, or miss the sphere centre

(0,0,3)

, radius 2?

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Sphere tangent to a plane means $p = ?$
2.
If $p < r$ , the plane cuts a circle of radius?
3.
If $p > r$ the plane and sphere?
4.
z-axis tangent to a sphere means distance from centre to z-axis = ?

From the bank · past-year question

Example 43D GeometryMODERATE

What is the length of the diameter of the sphere whose centre is at

(1,-2,3)

and which touches the plane

6x-3y+2z-4=0

[Q61 · Apr · 2020]

Touching a PLANE vs touching an AXIS

Tangent to a plane → use the point-to-plane distance for

p

. Tangent to the z-axis → use the distance from the centre to the z-axis,

\sqrt{x_c^2 + y_c^2}

. Don't mix the two distance formulas.

Drill 3 more on sphere and a plane — tangency and sections

Concept 5 of 6

Sphere and the coordinate axes

Intuition

Counting spheres of a fixed radius that touch all three coordinate axes is a symmetry puzzle: the centre must sit at distance

r

from each axis, which forces

|x_c| = |y_c| = |z_c|

with a sign choice per octant — giving a small finite count.

Definition

A sphere of radius $r$ touches all three axes when its centre is equidistant from each, with each distance equal to $r$ . Writing the centre as $(\pm a, \pm a, \pm a)$ , the distance from the centre to (say) the $z$ -axis is $\sqrt{a^2 + a^2} = a\sqrt2$ ; setting $a\sqrt2 = r$ gives $a = \tfrac{r}{\sqrt2}$ . Each independent sign choice gives a distinct sphere, so there are several such spheres (one per octant configuration).

Distance from a point to the z-axis

\text{dist to } z\text{-axis} = \sqrt{x_c^2 + y_c^2}

Worked example

A sphere of radius 6 touches all three coordinate axes and lies in the first octant. Find the coordinates of its centre.

Touching all three axes makes the centre equidistant from each, so write it as $(a, a, a)$ with $a > 0$ (first octant).
The distance from the centre to the $z$ -axis is $\sqrt{a^2 + a^2} = a\sqrt2$ , and this must equal the radius: $a\sqrt2 = 6$ .
Solve: $a = \dfrac{6}{\sqrt2} = 3\sqrt2$ .

Answer:Centre

= (3\sqrt2,\ 3\sqrt2,\ 3\sqrt2)

Practice this conceptself-check · 4 quick reps

Try it yourself

A sphere has centre

(3, 4, 7)

. What is its distance from the z-axis, and is the z-axis tangent if

r = 5

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Distance from $(x_c,y_c,z_c)$ to the z-axis?
2.
Spheres of radius r touching all three axes — how many?
3.
z-axis tangent ⇒ distance from centre to z-axis equals?
4.
Distance from $(6,8,1)$ to the z-axis?

From the bank · past-year question

Example 53D GeometryMODERATE

The number of spheres of radius

r

touching the coordinate axes is

[Q64 · Sep · 2021]

Drill 1 more on sphere and the coordinate axes

Concept 6 of 6

Locus problems with spheres

Intuition

The chapter's hardest sphere questions ask for the LOCUS of a moving point — the centre of a variable sphere, or the centroid of the triangle where a sphere cuts the axes. The method is always the same: write the moving point's coordinates in terms of a parameter, then eliminate the parameter to get a relation in

x, y, z

Definition

Set the moving point $(x, y, z)$ equal to the expression you're tracking (centre, centroid, etc.), express the free parameters from those equations, and substitute into the governing condition (the sphere passes through the origin, the plane through a fixed point, etc.). The resulting equation in $x, y, z$ — usually itself a sphere or plane — is the locus.

Worked example

A sphere of radius

r

through the origin cuts the axes at

A, B, C

. Find the locus of the centroid of triangle

ABC

Such a sphere is $x^2+y^2+z^2 - 2ux - 2vy - 2wz = 0$ (through origin → $d=0$ ); it meets the axes at $A(2u,0,0), B(0,2v,0), C(0,0,2w)$ .
Centroid $(x,y,z) = \left(\tfrac{2u}{3}, \tfrac{2v}{3}, \tfrac{2w}{3}\right)$ , so $u = \tfrac{3x}{2}$ etc.
Radius: $r = \sqrt{u^2+v^2+w^2}\Rightarrow u^2+v^2+w^2 = r^2$ .
Substitute: $\tfrac{9}{4}(x^2+y^2+z^2) = r^2$ , i.e. $x^2+y^2+z^2 = \tfrac{4r^2}{9}$ .

Answer:A sphere:

x^2 + y^2 + z^2 = \tfrac{4r^2}{9}

Practice this conceptself-check · 4 quick reps

Try it yourself

A point

P

moves so that its distance from the origin is always twice its distance from the point

(3, 0, 0)

. Find the locus of

P

Practice — Level 1 (4 reps)

Quick reps to lock in the method. Try each, then check.

1.
Locus method: write the point in terms of a parameter, then?
2.
A sphere through the origin has which constant zero?
3.
Sphere $x^2+y^2+z^2-2ux-2vy-2wz=0$ meets the x-axis at?
4.
The locus of a centre is often itself a?

From the bank · past-year question

Example 63D GeometryHARD

A variable plane passes through a fixed point

(a, b, c)

and cuts the axes in

A

B

and

C

respectively. The locus of the centre of the sphere

OABC

O

being the origin, is

[Q59 · Sep · 2017]

Drill 1 more on locus problems with spheres

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (4)

General equation, centre, and radius
Centre and radius of a sphere
$\text{centre} = (-u, -v, -w), \quad r = \sqrt{u^2 + v^2 + w^2 - d}$
Diameter form of a sphere
Diameter form
$(x-x_1)(x-x_2) + (y-y_1)(y-y_2) + (z-z_1)(z-z_2) = 0$
Sphere and a plane — tangency and sections
Tangency condition
$p = r \quad\text{(perpendicular distance from centre = radius)}$
Sphere and the coordinate axes
Distance from a point to the z-axis
$\text{dist to } z\text{-axis} = \sqrt{x_c^2 + y_c^2}$

Watch out for (1)

Touching a PLANE vs touching an AXIS
→ Sphere and a plane — tangency and sections

Mastery check — 5 interleaved questions

Try each one before clicking. Questions are interleaved across the concepts above, not grouped — interleaving sharpens transfer.

Example 13D GeometryHARD

For the following two (02) items: Suppose

S

is the sphere with the smallest radius that passes through the points

A(1,0,0)

B(0,1,0)

and

C(0,0,1)

What is the radius of

S

[Q53 · Sep · 2025]

Example 23D GeometryEASY

for the items that follow: The equation of the sphere S is

x^2+y^2+z^2-4x-6y-12z+k=0

What is the radius of the sphere passing through origin and concentric with the sphere S?

[Q63 · Apr · 2026]

Example 33D GeometryMODERATE

Consider the following for the items that follow: The plane

6x+ky+3z-12=0

where

k eq0

meets the coordinate axes at

A

B

and

C

respectively. The equation of the sphere passing through the origin and

A

B

C

x^2+y^2+z^2-2x-3y-4z=0

What is the value of

k

[Q91 · Apr · 2022]

Example 43D GeometryMODERATE

Consider the equation of a sphere

x^2 + y^2 + z^2 - 4x - 6y - 8z - 16 = 0

. Which of the following statements is/are correct? 1. z-axis is tangent to the sphere. 2. The centre of the sphere lies on the plane

x + y + z - 9 = 0

. Select the correct answer using the code given below:

[Q63 · Sep · 2022]

Example 53D GeometryHARD

A sphere of radius r through origin cuts axes at A,B,C. What is the locus of centroid of triangle ABC?

[Q60 · Apr · 2018]

Drill every past-year question on this subtopic

20 questions from the bank — paginated, with cart and Word-export support.

Drill the 20 questions