Sphere

At a Glance

• Frequency: 17 sub-parts across 12 of 13 years (2013, 2014, 2015, 2016, 2017, 2018, 2019, 2021, 2022, 2023, 2024, 2025)
• Priority tier: T1
• Marks (count): 10 (7), 12 (1), 13 (1), 15 (8)
• Average solve time: ~10 min
• Difficulty mix: medium 10, easy 5, hard 2
• Section: A | Dominant type: computation

Why This Chapter Matters

The sphere is the single most-tested topic in Analytic Geometry — it has appeared in 12 of the last 13 years (2013–2025), almost always as a 10- or 15-mark Section A question. The object never changes; only the condition imposed on it does. A small, fixed toolkit — the general equation, the tangency test, the pencil of spheres, and the orthogonality condition — disposes of every variant. Learn those four ideas well and you bank 10–15 near-certain marks each year.

Minimum Theory

The general sphere. Every sphere can be written $x^2+y^2+z^2+2ux+2vy+2wz+d=0,$ with centre $C=(-u,-v,-w)$ and radius $r=\sqrt{u^2+v^2+w^2-d}$. There are four unknowns $u,v,w,d$, so a sphere is pinned down by four independent conditions — four points, or any mix of point / tangency / orthogonality requirements.

Tangency and the pencil. A plane $\pi$ touches the sphere iff the perpendicular distance from the centre equals the radius, $\operatorname{dist}(C,\pi)=r$; the point of contact is the foot of that perpendicular. To construct a sphere through a given circle, use the pencil: if $S=0$ is a sphere and $L=0$ is a plane or second sphere meeting it in that circle, then every sphere through the circle is $S+\lambda L=0$ for some scalar $\lambda$, and one further condition fixes $\lambda$.

Orthogonality. Two spheres with coefficients $(u_i,v_i,w_i,d_i)$ cut orthogonally iff $2(u_1u_2+v_1v_2+w_1w_2)=d_1+d_2,$ equivalently $d^{\,2}=r_1^2+r_2^2$ along the line of centres. (Fuller derivations of the pencil and orthogonality conditions → appendix.)

Question Archetypes

Nine recurring patterns cover every sphere question in the corpus. Use this map to classify a question in seconds, then jump to the matching section.

pencil-family-construction (4 question(s); 2013, 2016, 2023, 2025)

Recognition Cues

• The question says “through the circle …” or “through the intersection of” a sphere and a plane (or two spheres).
• You are given two surfaces that define a circle, plus exactly one extra condition.
• That extra condition is one of: passes through the point …, is a great circle, is cut in a circle of radius …, or touches the plane ….

Solution Template

1. Name the circle’s surfaces. Identify $S=0$ (a sphere) and $L=0$ (a plane or a second sphere) whose intersection is the given circle.
2. Write the pencil $S+\lambda L=0$. By construction every member passes through the circle.
3. Read off the centre $C_\lambda$ (and radius$^2$, if the condition needs it) as functions of $\lambda$.
4. Impose the one extra condition to get a single equation in $\lambda$:
   • point on sphere → substitute the point;
   • great circle → $C_\lambda$ lies on the plane $L$;
   • given cut-radius → $r^2=R^2-d^2$ with $d=\operatorname{dist}(C_\lambda,\text{plane})$;
   • tangent to a plane → $\operatorname{dist}(C_\lambda,\text{plane})=R$.
5. Back-substitute $\lambda$, clear fractions, state the sphere(s). Expect two answers when the condition is quadratic in $\lambda$.

Worked Example(s)

2013 Paper 1, 2013-P1-Q1e (10 marks)

Sphere with intersection of given sphere and plane as great circle.

1 — Surfaces. The sphere $S$ has the segment from $(0,1,0)$ to $(3,-5,2)$ as a diameter, so the diameter form gives $S:\;x(x-3)+(y-1)(y+5)+z(z-2)=0\;\Longrightarrow\;x^2+y^2+z^2-3x+4y-2z-5=0,$ and the plane is $\pi:5x-2y+4z+7=0$.

2–3 — Pencil and centre. $S+\lambda\pi=0$ has centre $C_\lambda=\left(\tfrac{3-5\lambda}{2},\;\lambda-2,\;1-2\lambda\right).$

4 — Condition (great circle). The section plane must pass through the centre, i.e. $C_\lambda\in\pi$: $5\cdot\tfrac{3-5\lambda}{2}-2(\lambda-2)+4(1-2\lambda)+7=0\;\Longrightarrow\;45=45\lambda\;\Longrightarrow\;\lambda=1.$

5 — Result. $\boxed{\,x^2+y^2+z^2+2x+2y+2z+2=0\,}$ centre $(-1,-1,-1)$, radius $1$. Check: $5(-1)-2(-1)+4(-1)+7=0$, so the centre lies on $\pi$. ✓

2016 Paper 1, 2016-P1-Q1d (10 marks)

Find the sphere through the circle x^2+y^2=4, z=0 cut by x+2y+2z=0 in a circle of radius 3.

1 — Surfaces. The circle $x^2+y^2=4,\,z=0$ is the sphere $x^2+y^2+z^2-4=0$ met by the plane $z=0$.

2–3 — Pencil and centre. $x^2+y^2+z^2-4+\lambda z=0$; writing $\lambda=2f$ gives centre $C=(0,0,-f)$ and $R^2=f^2+4$.

4 — Condition (cut-radius $3$). Distance from $C$ to $x+2y+2z=0$ is $d=\dfrac{|-2f|}{\sqrt{1+4+4}}=\dfrac{2|f|}{3}$. The chord relation $r^2=R^2-d^2$ with $r=3$: $9=(f^2+4)-\tfrac{4f^2}{9}\;\Longrightarrow\;\tfrac{5f^2}{9}=5\;\Longrightarrow\;f=\pm3.$

5 — Result. Since $2f=\pm6$, $\boxed{\,x^2+y^2+z^2\pm 6z-4=0\,}$ two spheres, centres $(0,0,\mp3)$, radius $\sqrt{13}$.

2023 Paper 1, 2023-P1-Q3c (15 marks)

Sphere through circle x^2+y^2+z^2-4x-6y+2z-16=0, 3x+y+3z-4=0 in two cases (point on it; great circle).

Here $S\equiv x^2+y^2+z^2-4x-6y+2z-16$ and $L\equiv 3x+y+3z-4$; the pencil is $S+\lambda L=0$.

Case (i) — through $(1,0,-3)$. Substitute the point: $(1+0+9-4-0-6-16)+\lambda(3+0-9-4)=0\;\Longrightarrow\;-16-10\lambda=0\;\Longrightarrow\;\lambda=-\tfrac85.$ Clearing fractions ($\times 5$): $\boxed{\,5(x^2+y^2+z^2)-44x-38y-14z-48=0.\,}$

Case (ii) — given circle is a great circle. The centre must lie on $L$. With $C=\left(\tfrac{4-3\lambda}{2},\,\tfrac{6-\lambda}{2},\,-\tfrac{2+3\lambda}{2}\right)$, substitute into $3x+y+3z=4$: $\tfrac{12-19\lambda}{2}=4\;\Longrightarrow\;\lambda=\tfrac{4}{19},$ giving ($\times 19$): $\boxed{\,19(x^2+y^2+z^2)-64x-110y+50z-320=0.\,}$

2025 Paper 1, 2025-P1-Q3b (15 marks)

Find the spheres through the circle x^2+y^2+z^2-2x+2y+4z-3=0, 2x+y+z=4 and touching the plane 3x+4y=14.

With $S\equiv x^2+y^2+z^2-2x+2y+4z-3$ and $U\equiv 2x+y+z-4$, the pencil $S+\lambda U=0$ has centre $C=\left(1-\lambda,\;-\tfrac{2+\lambda}{2},\;-\tfrac{4+\lambda}{2}\right).$

Condition (tangent to $3x+4y-14=0$). The distance from $C$ to the plane simplifies cleanly: $\operatorname{dist}=\frac{\left|3(1-\lambda)+4\!\left(-\tfrac{2+\lambda}{2}\right)-14\right|}{5}=\frac{|-15-5\lambda|}{5}=|3+\lambda|.$ Setting $\operatorname{dist}^2=R^2$ and simplifying gives $\lambda^2-2\lambda=0$, so $\lambda=0$ or $\lambda=2$: $\boxed{\,x^2+y^2+z^2-2x+2y+4z-3=0\,}\quad(\lambda=0;\ \text{centre }(1,-1,-2),\,r=3)$ $\boxed{\,x^2+y^2+z^2+2x+4y+6z-11=0\,}\quad(\lambda=2;\ \text{centre }(-1,-2,-3),\,r=5)$ Check: $|3+0|=3=r$ and $|3+2|=5=r$. ✓ Note $\lambda=0$ simply recovers the original sphere $S$.

Common Traps

• “Great circle” = section plane through the centre. Enforce centre lies on the plane $L$ and verify it numerically; reading it as “the circle passes through a point” gives the wrong $\lambda$.
• Add only the term the circle’s plane allows. For a circle in $z=0$ the pencil is $x^2+y^2+z^2-4+\lambda z=0$; adding $x$- or $y$-terms slides the sphere off the circle.
• Use the perpendicular distance, properly normalised. In both $r^2=R^2-d^2$ and the tangency test, divide by $\sqrt{A^2+B^2+C^2}$ (e.g. $\sqrt{1+4+4}=3$).
• Expect two spheres when the condition is quadratic in $\lambda$ (cut-radius, tangency) — report both. Clearing fractions should leave all-integer coefficients, a useful correctness signal.

sphere-center-locus (3 question(s); 2017, 2021, 2022)

Recognition Cues

• A variable plane meets the coordinate axes at $A,B,C$, subject to a constraint (passes through a fixed point $(a,b,c)$, or the sphere has constant radius).
• A sphere is taken through $O,A,B,C$, and you must find the locus of its centre (or of the foot of the perpendicular from $O$ to plane $ABC$).

Solution Template

1. Intercept-form plane $\dfrac{x}{\alpha}+\dfrac{y}{\beta}+\dfrac{z}{\gamma}=1$, meeting the axes at $A(\alpha,0,0),B(0,\beta,0),C(0,0,\gamma)$. Write the given constraint:
   • plane through $(a,b,c)$ → $\dfrac{a}{\alpha}+\dfrac{b}{\beta}+\dfrac{c}{\gamma}=1$;
   • constant radius $r$ → $\alpha^2+\beta^2+\gamma^2=4r^2$.
2. Sphere through $O,A,B,C$. Through $O$ kills the constant term ($d=0$); through $A,B,C$ gives $u=-\tfrac\alpha2,\,v=-\tfrac\beta2,\,w=-\tfrac\gamma2$, so the sphere is $x^2+y^2+z^2-\alpha x-\beta y-\gamma z=0$ with centre $(\tfrac\alpha2,\tfrac\beta2,\tfrac\gamma2)$.
3. Introduce the locus point $(X,Y,Z)$ (the centre, or the foot of the perpendicular) and express $\alpha,\beta,\gamma$ in terms of it (e.g. $\alpha=2X$).
4. Eliminate $\alpha,\beta,\gamma$ with the constraint; replace $(X,Y,Z)\to(x,y,z)$ for the locus.

Worked Example(s)

2017 Paper 1, 2017-P1-Q2b (15 marks)

Plane through fixed (a,b,c) cuts axes at A,B,C; find locus of centre of sphere through O,A,B,C.

1 — Plane + constraint. $\dfrac{x}{p}+\dfrac{y}{q}+\dfrac{z}{r}=1$ through $(a,b,c)$ gives $\dfrac{a}{p}+\dfrac{b}{q}+\dfrac{c}{r}=1.$

2 — Sphere through $O,A,B,C$ is $x^2+y^2+z^2-px-qy-rz=0$, centre $\left(\tfrac p2,\tfrac q2,\tfrac r2\right)$.

3 — Locus point. Set the centre $=(x,y,z)$, so $p=2x,\ q=2y,\ r=2z$.

4 — Eliminate. Substituting into the constraint: $\dfrac{a}{2x}+\dfrac{b}{2y}+\dfrac{c}{2z}=1\;\Longrightarrow\;\boxed{\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=2,}$ equivalently $2xyz=a\,yz+b\,zx+c\,xy$. (The $1$ becomes a $2$ precisely because the centre is at half the intercepts.)

2021 Paper 1, 2021-P1-Q3b (15 marks)

Sphere radius r through O cuts axes at A,B,C; locus of foot of perp from O to plane ABC.

This is the constant-radius / foot-of-perpendicular variant. Sphere through $O$ and $A(\alpha,0,0),B,C$ has centre $(\tfrac\alpha2,\tfrac\beta2,\tfrac\gamma2)$ and radius $\tfrac12\sqrt{\alpha^2+\beta^2+\gamma^2}=r$, so the constraint is $\alpha^2+\beta^2+\gamma^2=4r^2$.

Foot of perpendicular $P=(X,Y,Z)$ from $O$ to plane $ABC$ lies along the normal $(1/\alpha,1/\beta,1/\gamma)$; substituting $P=t(1/\alpha,1/\beta,1/\gamma)$ into the plane gives $t=\big(\sum 1/\alpha^2\big)^{-1}$, and one finds $t=X^2+Y^2+Z^2$ with $\alpha=t/X$ etc.

Eliminate via $\alpha^2+\beta^2+\gamma^2=4r^2$: $\boxed{\,(x^2+y^2+z^2)^2\!\left(\dfrac1{x^2}+\dfrac1{y^2}+\dfrac1{z^2}\right)=4r^2.\,}$ (Check with the symmetric sphere $\alpha=\beta=\gamma=2r/\sqrt3$: the foot is $X=Y=Z=2r/(3\sqrt3)$, and the locus reduces to $27X^2=4r^2$. ✓)

2022 Paper 1, 2022-P1-Q1e (10 marks)

Variable plane through (a,b,c) meets axes at A,B,C; find locus of centre of sphere through O,A,B,C.

Identical setup to the 2017 example (a plane through a fixed point $(a,b,c)$; locus of the centre of the sphere through $O,A,B,C$). Centre $=\left(\tfrac\alpha2,\tfrac\beta2,\tfrac\gamma2\right)$, set it to $(x,y,z)$ so $\alpha=2x$ etc., and the through-point condition $\dfrac{a}{\alpha}+\dfrac{b}{\beta}+\dfrac{c}{\gamma}=1$ becomes $\boxed{\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=2.}$ That this 10-mark question (2022) and the 15-mark question (2017) have the same answer shows how stable this archetype is — learn it once.

Common Traps

• The centre is at half the intercepts ($\alpha=2x$). Forgetting the factor of 2 turns the $1$ in the constraint into a wrong constant — the right locus is $\tfrac ax+\tfrac by+\tfrac cz=2$, not $=1$.
• “Through $O$” kills the constant term ($d=0$); the linear coefficients then come from passing through $A,B,C$ — derive $u=-\tfrac\alpha2$, don’t guess the sign.
• “Constant radius $r$” encodes $\alpha^2+\beta^2+\gamma^2=4r^2$ — that is the constraint you eliminate with.
• Keep the locus in centre-coordinates $(x,y,z)$; do not leave the intercepts $\alpha,\beta,\gamma$ in the final answer.

tangent-plane-contact (3 question(s); 2014, 2015, 2017)

Recognition Cues

• The words “touches”, “tangent plane”, or “point of contact”.
• Asked to find a parameter (e.g. $a$) that makes a plane tangent.
• “tangent plane parallel to [a given plane]” — a variant where the radius to the contact point is parallel to that plane’s normal.

Solution Template

1. Standard form. Complete squares to read off centre $C$ and radius $R$.
2. Tangency test. A plane touches iff $\operatorname{dist}(C,\text{plane})=R$. To find an unknown parameter, set $\operatorname{dist}=R$ and solve (often a quadratic).
3. Point of contact = foot of perpendicular from $C$: $\text{contact}=C-\frac{s}{|\mathbf n|^{2}}\,\mathbf n,\qquad s=\text{(plane expression evaluated at }C),$ keeping the sign of $s$ so you step to the correct side.
4. Parallel-plane variant. If the tangent plane must be parallel to a plane with normal $\mathbf n$, the contact points are $C\pm R\,\hat{\mathbf n}$ (the two ends of the diameter along $\mathbf n$).
5. Verify the contact lies on both the plane and the sphere, and $|CP|=R$.

Worked Example(s)

2014 Paper 1, 2014-P1-Q4a-i (10 marks)

Find points on sphere x²+y²+z²-4x+2y=4 where tangent plane parallel to 2x-y+2z=1.

This is the parallel-plane variant (template step 4).

1 — Standard form. $x^2+y^2+z^2-4x+2y=4\Rightarrow(x-2)^2+(y+1)^2+z^2=9$, so $C=(2,-1,0)$, $R=3$.

4 — Radius parallel to the given normal. The given plane $2x-y+2z=1$ has normal $\mathbf n=(2,-1,2)$, $|\mathbf n|=3=R$. The contact points are $C\pm\mathbf n=(2,-1,0)\pm(2,-1,2),$ $\boxed{\,P_1=(4,-2,2),\quad P_2=(0,0,-2).\,}$ They are diametrically opposite, as they must be — the two parallel tangent planes sit on opposite sides of the sphere. (Both satisfy the sphere equation: $4+1+4=9$.)

2015 Paper 1, 2015-P1-Q1e (10 marks)

For positive a, find when plane ax-2y+z+12=0 touches sphere x^2+y^2+z^2-2x-4y+2z-3=0; find point of contact.

1 — Standard form. $x^2+y^2+z^2-2x-4y+2z-3=0\Rightarrow(x-1)^2+(y-2)^2+(z+1)^2=9$, so $C=(1,2,-1)$, $R=3$.

2 — Tangency fixes $a$. With plane $ax-2y+z+12=0$, $\operatorname{dist}(C,\text{plane})=\frac{|a-4-1+12|}{\sqrt{a^2+4+1}}=\frac{|a+7|}{\sqrt{a^2+5}}=3.$ Squaring: $(a+7)^2=9(a^2+5)\Rightarrow 4a^2-7a-2=0\Rightarrow a=\dfrac{7\pm9}{8}$, i.e. $a=2$ or $a=-\tfrac14$. The positive value is $\boxed{a=2}$.

3 — Point of contact (now the plane is $2x-2y+z+12=0$, $\mathbf n=(2,-2,1)$, $|\mathbf n|^2=9$). Here $s=2(1)-2(2)+(-1)+12=9$, so $\text{contact}=C-\frac{9}{9}(2,-2,1)=(1,2,-1)-(2,-2,1)=\boxed{(-1,4,-2)}.$

5 — Verify. On the plane: $-2-8-2+12=0$ ✓; $|CP|^2=4+4+1=9$ ✓.

2017 Paper 1, 2017-P1-Q2c (10 marks)

Show the plane 2x-2y+z+12=0 touches a given sphere; find the point of contact.

This is the pure tangency test on the same sphere as the previous example, $C=(1,2,-1)$, $R=3$.

2 — Show tangency. For $2x-2y+z+12=0$, $\mathbf n=(2,-2,1)$, $|\mathbf n|=3$: $\operatorname{dist}(C,\text{plane})=\frac{|2(1)-2(2)+(-1)+12|}{3}=\frac{9}{3}=3=R,$ so the plane touches the sphere.

3 — Contact. As above, $\text{contact}=C-\dfrac{9}{9}(2,-2,1)=\boxed{(-1,4,-2)}$, lying on both surfaces. ✓

Common Traps

• Keep the sign of $s$ when stepping from the centre to the contact point — the absolute value lands you on the wrong side of the sphere.
• Keep $|\mathbf n|^{2}$ in the denominator of the foot-of-perpendicular formula; that $|\mathbf n|=R=3$ in these examples is a coincidence, not a rule.
• A tangency quadratic has two roots — read the question for which one is wanted (here, $a>0$).
• Verify the contact point on both the plane and the sphere. It is the cleanest confirmation; for the parallel-plane variant, remember the two contact points are diametrically opposite.

sphere-through-points (2 question(s); 2018, 2019)

Recognition Cues

• A sphere is required through given points — typically four (a sphere has four degrees of freedom).
• A “circle through $A,B,C$” or “circle circumscribing triangle $ABC$” in space — recognise it as (a sphere through $A,B,C$) $\cap$ (their plane).

Solution Template

1. General sphere $x^2+y^2+z^2+2ux+2vy+2wz+d=0$ (four unknowns $u,v,w,d$).
2. Substitute the easiest point first. If the origin is among the points, $d=0$ immediately and the rest of the algebra is much lighter.
3. Substitute the remaining points to get a linear system in $u,v,w$ (and $d$); solve it.
4. Clear fractions and state the sphere. Circle variant: a circle in space needs two equations — present the sphere and the plane.
5. Verify every given point satisfies the final equation.

Worked Example(s)

2018 Paper 1, 2018-P1-Q3d (12 marks)

Find the sphere through (0,0,0),(0,1,-1),(-1,2,0),(1,2,3).

1–2 — Origin first. $(0,0,0)$ gives $d=0$, so $x^2+y^2+z^2+2ux+2vy+2wz=0$.

3 — Remaining points. $(0,1,-1):\ 2+2v-2w=0\Rightarrow v-w=-1;$ $(-1,2,0):\ 5-2u+4v=0\Rightarrow -2u+4v=-5;$ $(1,2,3):\ 14+2u+4v+6w=0\Rightarrow u+2v+3w=-7.$ Solving: $w=v+1$, then $u+5v=-10$ and $u=2v+\tfrac52$ give $v=-\tfrac{25}{14},\ u=-\tfrac{15}{14},\ w=-\tfrac{11}{14}$.

4 — Sphere. Clearing the sevenths, $\boxed{\,7(x^2+y^2+z^2)-15x-25y-11z=0.\,}$

5 — Verify. $(1,2,3)$: $7(14)-15-50-33=98-98=0$ ✓; the other three check similarly. Centre $\left(\tfrac{15}{14},\tfrac{25}{14},\tfrac{11}{14}\right)$, radius$^2=\tfrac{971}{196}>0$ — a genuine sphere.

2019 Paper 1, 2019-P1-Q2c-i (10 marks)

Plane x+2y+3z=12 meets the axes in A,B,C; find the circle circumscribing triangle ABC.

This is the circle variant. The plane $x+2y+3z=12$ meets the axes at $A(12,0,0),\,B(0,6,0),\,C(0,0,4)$. The circumscribing circle of $\triangle ABC$ is (a sphere through $A,B,C$) $\cap$ (the plane); take the sphere through $O,A,B,C$.

Sphere through $O,A,B,C$ ($d=0$): $A\Rightarrow 144+24u=0\Rightarrow u=-6$; $B\Rightarrow v=-3$; $C\Rightarrow w=-2$, so $x^2+y^2+z^2-12x-6y-4z=0.$

The circle is the pair of equations: $\boxed{\,x^2+y^2+z^2-12x-6y-4z=0,\qquad x+2y+3z=12.\,}$ (Sphere centre $(6,3,2)$, $R=7$; distance to the plane $6/\sqrt{14}<7$, so the section is a real circle.)

Common Traps

• Substitute the origin first when it is one of the points — it kills $d$ and lightens every later equation. Keep coefficients as exact fractions, then clear to a tidy integer form.
• A circle in space needs two equations — a sphere and a plane. Reporting only the sphere is a marks-losing omission; the origin-passing sphere is just the cleanest choice of sphere through the three points.
• Always verify all given points satisfy the final equation — the cheapest insurance against an arithmetic slip in the linear system.

extremal-point (1 question(s); 2015)

Recognition Cues

• Asked for the point of the sphere at maximum (or minimum) distance from a given external point.

Solution Template

1. Identify centre $C$, radius $r$, and the external point $Q$.
2. The extreme points lie on the line $QC$. This is pure geometry — no Lagrange multipliers needed.
3. Nearest $=C+r\,\widehat{CQ}$ (toward $Q$), at distance $|CQ|-r$; farthest $=C-r\,\widehat{CQ}$ (opposite side of $C$ from $Q$), at distance $|CQ|+r$.

Worked Example(s)

2015 Paper 1, 2015-P1-Q3b (13 marks)

Point of sphere x^2+y^2+z^2=1 at max distance from (2,1,3).

1 — Data. $x^2+y^2+z^2=1$ has $C=O$, $r=1$; external point $Q=(2,1,3)$ with $|CQ|=\sqrt{4+1+9}=\sqrt{14}>1$.

2–3 — Farthest point. It lies on line $QC$, on the far side of $C$ from $Q$, i.e. in direction $-\widehat{CQ}=-\tfrac{1}{\sqrt{14}}(2,1,3)$: $\boxed{\,Q^\ast=\tfrac{1}{\sqrt{14}}(-2,-1,-3),\qquad \text{max distance}=|CQ|+r=\sqrt{14}+1.\,}$ The answer points opposite to $(2,1,3)$ — the quickest sanity check that it is the farthest, not the nearest, point. ($|Q^\ast|=1$, so it is on the sphere.)

Common Traps

• Reach for geometry, not Lagrange multipliers. The extreme points are simply where line $QC$ meets the sphere — instant, and impossible to get wrong.
• Farthest is the opposite direction. The farthest point lies on the far side of the centre from $Q$ (direction $-\widehat{CQ}$); the nearest is on the near side. Mixing them up flips the sign of the answer.

inscribed-tangent-sphere (1 question(s); 2022)

Recognition Cues

• The smallest sphere that touches two (skew) lines.

Solution Template

1. Read off a point and direction on each line: $P_1,\vec d_1$ and $P_2,\vec d_2$.
2. Find the feet of the common perpendicular $A=P_1+s\vec d_1$, $B=P_2+t\vec d_2$ by solving $\vec{AB}\cdot\vec d_1=0$ and $\vec{AB}\cdot\vec d_2=0$.
3. Centre $=$ midpoint of $AB$; radius $=\tfrac12|AB|=\tfrac12\times$(shortest distance).
4. Write the sphere and verify its distance to each line equals the radius.

Worked Example(s)

2022 Paper 1, 2022-P1-Q3c (15 marks)

Sphere of smallest radius tangent to two given skew lines.

1 — Data. $L_1:\ P_1=(3,8,3),\ \vec d_1=(3,-1,1)$; $L_2:\ P_2=(-3,-7,6),\ \vec d_2=(-3,2,4)$, so $\vec{P_1P_2}=(-6,-15,3)$.

2 — Feet of the common perpendicular. Solving $\vec{AB}\cdot\vec d_1=0,\ \vec{AB}\cdot\vec d_2=0$ here gives $s=t=0$, because $\vec{P_1P_2}\cdot\vec d_1=\vec{P_1P_2}\cdot\vec d_2=0$ already. So $P_1,P_2$ are themselves the feet — a pleasant shortcut.

3 — Centre and radius. Centre $=\tfrac12(P_1+P_2)=\left(0,\tfrac12,\tfrac92\right)$; radius $=\tfrac12|\vec{P_1P_2}|=\tfrac12\sqrt{270}=\tfrac{3\sqrt{30}}2$.

4 — Sphere. $\boxed{\,x^2+\left(y-\tfrac12\right)^2+\left(z-\tfrac92\right)^2=\tfrac{135}{2}\,}\quad\bigl(\text{i.e. }x^2+y^2+z^2-y-9z=47\bigr).$ The centre’s distance to each line works out to $\tfrac{3\sqrt{30}}2=r$. ✓

Common Traps

• Always test for the shortcut. If $\vec{P_1P_2}\cdot\vec d_1=\vec{P_1P_2}\cdot\vec d_2=0$ (equivalently $\vec{P_1P_2}\parallel\vec d_1\times\vec d_2$), the given points are already the feet of the common perpendicular and $s=t=0$ — saving the whole linear solve.
• Smallest sphere $\Rightarrow$ diameter $=$ shortest distance. The centre is the midpoint of the common-perpendicular segment and the radius is half it; don’t use the full distance.

orthogonal-cutting (1 question(s); 2024)

Recognition Cues

• The required sphere must cut a given sphere orthogonally, usually combined with another condition (often touches a plane at a point).

Solution Template

1. General sphere $S$ with unknowns $(u,v,w,d)$; each condition gives one equation.
2. “Touches a plane $\pi$ at $P_0$” gives two: (a) $P_0$ lies on $S$; (b) the centre lies on the normal through $P_0$, so parametrise $C=P_0+\mu\,\vec n$ (cleanest for signs).
3. “Cuts $S_2$ orthogonally” gives $2(uu_2+vv_2+ww_2)=d+d_2$.
4. Solve for the parameter(s), recover $(u,v,w,d)$, and state the sphere.

Worked Example(s)

2024 Paper 1, 2024-P1-Q4c (15 marks)

Find sphere touching plane 3x+2y-z+2=0 at (1,-2,1) and cutting orthogonally x^2+y^2+z^2-4x+6y+4=0.

Touches $3x+2y-z+2=0$ at $P_0=(1,-2,1)$ and cuts $S_2:x^2+y^2+z^2-4x+6y+4=0$ orthogonally.

2(b) — Centre on the normal. With $\vec n=(3,2,-1)$, set $C=P_0+\mu\vec n$, giving $u=-1-3\mu,\ v=2-2\mu,\ w=-1+\mu$.

2(a) — $P_0$ on the sphere gives $d=6-4\mu$.

3 — Orthogonality with $S_2$ ($u_2=-2,v_2=3,w_2=0,d_2=4$): $-4u+6v=d+4$. Substituting, $4+12\mu+12-12\mu=10-4\mu\;\Longrightarrow\;16=10-4\mu\;\Longrightarrow\;\mu=-\tfrac32.$

4 — Recover and state. $u=\tfrac72,\ v=5,\ w=-\tfrac52,\ d=12$: $\boxed{\,x^2+y^2+z^2+7x+10y-5z+12=0.\,}$ Checks: $P_0$ on it ($1+4+1+7-20-5+12=0$); $\vec{CP_0}=\tfrac32(3,2,-1)\parallel\vec n$; $2(uu_2+vv_2+ww_2)=16=d+d_2$. ✓

Common Traps

• Parametrise the centre as $C=P_0+\mu\vec n$. It tracks signs far more reliably than writing $\vec{CP_0}\parallel\vec n$ with a separate scalar.
• Memorise the orthogonality condition $2(uu'+vv'+ww')=d+d'$ — it is just $|CC'|^2=R^2+R'^2$ expanded, and it is the load-bearing equation here.

tangency-existence (1 question(s); 2025)

Recognition Cues

• Asked to decide whether any plane through a given line can touch the sphere (often phrased “show that no tangent plane through the line exists”).

Solution Template

1. Sphere → centre $C$, radius $r$.
2. Write the line as the intersection of two planes $\pi_1,\pi_2$.
3. Pencil of planes through the line: $\pi_1+k\,\pi_2=0$.
4. Impose tangency $\operatorname{dist}(C,\text{plane})=r$ → a quadratic in $k$. Real roots ⇒ tangent planes exist; negative discriminant ⇒ none. (Also check the excluded $k\to\infty$ plane $\pi_2$.)
5. Cross-check geometrically: if $\operatorname{dist}(C,\text{line})<r$ the line is a secant, so every plane through it cuts the sphere and none can be tangent.

Worked Example(s)

2025 Paper 1, 2025-P1-Q4a (15 marks)

Show there is no tangent plane to the sphere x^2+y^2+z^2-4x+2y-4z+4=0 through the line (x+6)/2=y+3=z+1.

1 — Sphere. $x^2+y^2+z^2-4x+2y-4z+4=0$: centre $C=(2,-1,2)$, $r^2=4+1+4-4=5$.

2 — Line as two planes. $\dfrac{x+6}{2}=y+3=z+1$ gives $\pi_1:x-2y=0$ and $\pi_2:y-z+2=0$.

3 — Pencil. $\pi_1+k\pi_2=0:\ x+(k-2)y-kz+2k=0$.

4 — Tangency. Setting $\operatorname{dist}(C,\text{plane})^2=r^2$: $(4-k)^2=5\bigl(1+(k-2)^2+k^2\bigr)\;\Longrightarrow\;9k^2-12k+9=0\;\Longrightarrow\;3k^2-4k+3=0.$ Discriminant $=(-4)^2-4\cdot3\cdot3=-20<0$: no real $k$, so no plane of the pencil is tangent (and $\pi_2$ alone is at distance $1/\sqrt2\neq r$).

5 — Cross-check. $\operatorname{dist}(C,\text{line})^2=\tfrac72<5=r^2$, so the line is a secant — it pierces the sphere, confirming no plane through it can touch. $\boxed{\,\text{Discriminant }-20<0\ \Rightarrow\ \text{no tangent plane through the line exists.}\,}$

Common Traps

• The discriminant is the answer. “No tangent plane exists” $\Leftrightarrow$ the tangency quadratic in $k$ has negative discriminant; state that explicitly rather than just failing to find a root.
• Don’t forget the excluded plane. The pencil $\pi_1+k\pi_2$ omits $\pi_2$ itself (the $k\to\infty$ member); check it separately.
• Back it with the secant cross-check ($\operatorname{dist}(C,\text{line})<r$): a one-line geometric confirmation that graders reward.

tangent-lines-from-point (1 question(s); 2013)

Recognition Cues

• Tangent lines (a tangent cone) drawn from an external point to the sphere — especially three mutually perpendicular tangent lines (the director sphere).

Solution Template

1. Tangent-cone semi-angle. From an external point $P$, the tangent lines to a sphere (centre $O$, radius $r$) form a cone of semi-angle $\alpha$ with $\sin\alpha=r/|OP|$.
2. Three mutually perpendicular tangent lines. Each makes angle $\alpha$ with $OP$, so $(\hat n\cdot\hat u_i)^2=\cos^2\alpha$; orthonormality gives $\sum_i(\hat n\cdot\hat u_i)^2=1$, hence $3\cos^2\alpha=1$, i.e. $\cos^2\alpha=\tfrac13$.
3. Combine with $\sin\alpha=r/|OP|$ to get the locus of $P$ (the director sphere).

Worked Example(s)

2013 Paper 1, 2013-P1-Q4a (15 marks)

Three mutually perpendicular tangent lines from any point on sphere 2(x²+y²+z²)=3r² to sphere r.

Show that three mutually perpendicular tangent lines to $S:x^2+y^2+z^2=r^2$ can be drawn from any point on $2(x^2+y^2+z^2)=3r^2$.

1–2 — Angle condition. A point $P$ on $2(x^2+y^2+z^2)=3r^2$ has $|OP|^2=\tfrac{3r^2}{2}$, so the tangent cone has $\sin^2\alpha=\dfrac{r^2}{|OP|^2}=\dfrac{2}{3}$, hence $\cos^2\alpha=\tfrac13$.

3 — Orthonormal frame exists. Three mutually perpendicular tangent directions $\hat u,\hat v,\hat w$ would need $(\hat n\cdot\hat u)^2=(\hat n\cdot\hat v)^2=(\hat n\cdot\hat w)^2=\cos^2\alpha$, and orthonormality forces their sum to be $1$: $3\cos^2\alpha=1\Rightarrow\cos^2\alpha=\tfrac13$ — exactly the value above. So such a frame exists for every $P$ on the given sphere. $\boxed{\,\text{Three mutually perpendicular tangent lines exist from every point of }2(x^2+y^2+z^2)=3r^2.\,}$ (This locus, $|OP|=r\sqrt{3/2}$, is the director sphere for perpendicular tangent lines.)

Common Traps

• Lines, not planes. The director sphere for perpendicular tangent lines is $|OP|=r\sqrt{3/2}$; the perpendicular tangent planes result is the different $|OP|=r\sqrt3$. Don’t conflate them.
• The sum-of-squared-cosines identity is the whole trick. $\sum_i(\hat n\cdot\hat u_i)^2=1$ for any orthonormal frame turns “three perpendicular tangents” into the single equation $3\cos^2\alpha=1$.

Marks-Aware Writing

A correct final equation alone rarely earns full marks — examiners reward a visible, justified method.

• Set up explicitly. Write the general sphere $x^2+y^2+z^2+2ux+2vy+2wz+d=0$ (or the pencil $S+\lambda L=0$) and name the centre/radius before imposing anything. Don’t jump straight to the answer.
• Name each condition as you use it: “tangency $\Rightarrow \operatorname{dist}(C,\pi)=r$”, “great circle $\Rightarrow$ centre on the plane”, “orthogonal $\Rightarrow 2(uu'+vv'+ww')=d+d'$”. One labelled line per condition is worth easy marks.
• Show the elimination. A 15-mark answer is expected to carry the full algebra (solving the system / clearing fractions); a 10-mark answer should still show the key steps and the result, not just the final line.
• Box the equation and verify cheaply. A one-line check — a defining point satisfies the sphere, or $\operatorname{dist}=r$ — is reliably rewarded.
• Report every solution. Cut-radius and tangency conditions are quadratic, so give both spheres; loci should be left in clean $(x,y,z)$ form.

Practice Set

All 17 primary instances are worked above, so these are questions where the sphere plays a supporting role — good practice at deploying the toolkit inside a larger problem.

• 2024-P1-Q1e (10 m) — right circular cylinder through a circle: the circle is a sphere $\cap$ plane; the cylinder’s axis runs along the plane normal.
• 2023-P1-Q2c-ii (10 m) — cone from $O$ to circle $ABC$: build the auxiliary sphere $x^2+y^2+z^2=ax+by+cz$ through $O,A,B,C$, then take rays from $O$.
• 2023-P1-Q4c (15 m) — image of a line in a plane: the reflection uses the same foot-of-perpendicular step as tangent-contact.
• 2013-P1-Q4b (15 m) — cone through a base circle in $z=0$: impose the rectangular-hyperbola section condition (coeff sum $=0$).
• 2015-P1-Q6c (12 m) — orthogonal surfaces at a point: the same orthogonality idea via $\nabla F_1\cdot\nabla F_2=0$.
• 2019-P1-Q2c-ii (10 m) — enveloping cone cut by a plane: form $SS_1=T^2$, set $z=0$, apply coeff$(x^2)+$coeff$(y^2)=0$.