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UPSC 2026 Maths Optional Paper 1 Q1e — Step-by-Step Solution 10 marks · Section A
Sphere · Analytic Geometry · asked 18× in 14 yrs · Read the full method →
Question
Obtain the equation of the sphere having its centre on the line 5 y + 2 z = 0 = 2 x − 3 y 5y + 2z = 0 = 2x - 3y 5 y + 2 z = 0 = 2 x − 3 y ( 0 , − 2 , − 4 ) (0, -2, -4) ( 0 , − 2 , − 4 ) ( 2 , − 1 , − 1 ) (2, -1, -1) ( 2 , − 1 , − 1 )
Technique
The line is the intersection of two planes through the origin — parametrise it, so the centre carries a single unknown. Impose equidistance from the two given points (equivalently, the centre lies on the perpendicular bisector plane of the chord) to pin the parameter; the radius then follows, and the sphere is written in expanded form.
Solution
Step 1 — Parametrise the line carrying the centre.
The symbol 5 y + 2 z = 0 = 2 x − 3 y 5y+2z=0=2x-3y 5 y + 2 z = 0 = 2 x − 3 y
π 1 : 5 y + 2 z = 0 , π 2 : 2 x − 3 y = 0. \pi_1:\;5y+2z=0,\qquad \pi_2:\;2x-3y=0. π 1 : 5 y + 2 z = 0 , π 2 : 2 x − 3 y = 0. Both have zero constant term, so both pass through the origin; hence the line passes through O = ( 0 , 0 , 0 ) O=(0,0,0) O = ( 0 , 0 , 0 )
Its direction is perpendicular to both normals n 1 = ( 0 , 5 , 2 ) \mathbf n_1=(0,5,2) n 1 = ( 0 , 5 , 2 ) n 2 = ( 2 , − 3 , 0 ) \mathbf n_2=(2,-3,0) n 2 = ( 2 , − 3 , 0 )
n 1 × n 2 = ∣ i ^ j ^ k ^ 0 5 2 2 − 3 0 ∣ = i ^ ( 5 ⋅ 0 − 2 ⋅ ( − 3 ) ) − j ^ ( 0 ⋅ 0 − 2 ⋅ 2 ) + k ^ ( 0 ⋅ ( − 3 ) − 5 ⋅ 2 ) = ( 6 , 4 , − 10 ) = 2 ( 3 , 2 , − 5 ) . \mathbf n_1\times\mathbf n_2=\begin{vmatrix}\hat i & \hat j & \hat k\\ 0 & 5 & 2\\ 2 & -3 & 0\end{vmatrix} = \hat i(5\cdot0-2\cdot(-3)) - \hat j(0\cdot0-2\cdot2) + \hat k(0\cdot(-3)-5\cdot2) = (6,\,4,\,-10) = 2(3,2,-5). n 1 × n 2 = i ^ 0 2 j ^ 5 − 3 k ^ 2 0 = i ^ ( 5 ⋅ 0 − 2 ⋅ ( − 3 )) − j ^ ( 0 ⋅ 0 − 2 ⋅ 2 ) + k ^ ( 0 ⋅ ( − 3 ) − 5 ⋅ 2 ) = ( 6 , 4 , − 10 ) = 2 ( 3 , 2 , − 5 ) . So the line is x 3 = y 2 = z − 5 ( = s ) \dfrac{x}{3}=\dfrac{y}{2}=\dfrac{z}{-5}\;(=s) 3 x = 2 y = − 5 z ( = s )
C = ( 3 s , 2 s , − 5 s ) , s ∈ R . C=(3s,\;2s,\;-5s),\qquad s\in\mathbb{R}. C = ( 3 s , 2 s , − 5 s ) , s ∈ R . (Check: 5 ( 2 s ) + 2 ( − 5 s ) = 10 s − 10 s = 0 5(2s)+2(-5s)=10s-10s=0 5 ( 2 s ) + 2 ( − 5 s ) = 10 s − 10 s = 0 2 ( 3 s ) − 3 ( 2 s ) = 6 s − 6 s = 0 2(3s)-3(2s)=6s-6s=0 2 ( 3 s ) − 3 ( 2 s ) = 6 s − 6 s = 0
Step 2 — Impose that both given points lie on the sphere.
Let A = ( 0 , − 2 , − 4 ) A=(0,-2,-4) A = ( 0 , − 2 , − 4 ) B = ( 2 , − 1 , − 1 ) B=(2,-1,-1) B = ( 2 , − 1 , − 1 ) A , B A,B A , B ∣ C A ∣ = ∣ C B ∣ = |CA|=|CB|= ∣ C A ∣ = ∣ C B ∣ =
∣ C A ∣ 2 = ( 3 s − 0 ) 2 + ( 2 s + 2 ) 2 + ( − 5 s + 4 ) 2 = 9 s 2 + ( 4 s 2 + 8 s + 4 ) + ( 25 s 2 − 40 s + 16 ) = 38 s 2 − 32 s + 20 , |CA|^{2}=(3s-0)^{2}+(2s+2)^{2}+(-5s+4)^{2}=9s^{2}+\left(4s^{2}+8s+4\right)+\left(25s^{2}-40s+16\right)=38s^{2}-32s+20, ∣ C A ∣ 2 = ( 3 s − 0 ) 2 + ( 2 s + 2 ) 2 + ( − 5 s + 4 ) 2 = 9 s 2 + ( 4 s 2 + 8 s + 4 ) + ( 25 s 2 − 40 s + 16 ) = 38 s 2 − 32 s + 20 , ∣ C B ∣ 2 = ( 3 s − 2 ) 2 + ( 2 s + 1 ) 2 + ( − 5 s + 1 ) 2 = ( 9 s 2 − 12 s + 4 ) + ( 4 s 2 + 4 s + 1 ) + ( 25 s 2 − 10 s + 1 ) = 38 s 2 − 18 s + 6. |CB|^{2}=(3s-2)^{2}+(2s+1)^{2}+(-5s+1)^{2}=\left(9s^{2}-12s+4\right)+\left(4s^{2}+4s+1\right)+\left(25s^{2}-10s+1\right)=38s^{2}-18s+6. ∣ C B ∣ 2 = ( 3 s − 2 ) 2 + ( 2 s + 1 ) 2 + ( − 5 s + 1 ) 2 = ( 9 s 2 − 12 s + 4 ) + ( 4 s 2 + 4 s + 1 ) + ( 25 s 2 − 10 s + 1 ) = 38 s 2 − 18 s + 6. Step 3 — Solve for s s s
Set ∣ C A ∣ 2 = ∣ C B ∣ 2 |CA|^{2}=|CB|^{2} ∣ C A ∣ 2 = ∣ C B ∣ 2 38 s 2 38s^{2} 38 s 2
− 32 s + 20 = − 18 s + 6 ⟹ 14 = 14 s ⟹ s = 1. -32s+20=-18s+6\;\Longrightarrow\;14=14s\;\Longrightarrow\;s=1. − 32 s + 20 = − 18 s + 6 ⟹ 14 = 14 s ⟹ s = 1. ∴ C = ( 3 , 2 , − 5 ) . \therefore\quad C=(3,\,2,\,-5). ∴ C = ( 3 , 2 , − 5 ) . Independent cross-check (perpendicular bisector plane). Midpoint of A B AB A B M = ( 1 , − 3 2 , − 5 2 ) M=\left(1,-\tfrac32,-\tfrac52\right) M = ( 1 , − 2 3 , − 2 5 ) A B → = ( 2 , 1 , 3 ) \overrightarrow{AB}=(2,1,3) A B = ( 2 , 1 , 3 ) A A A B B B
2 ( x − 1 ) + 1 ( y + 3 2 ) + 3 ( z + 5 2 ) = 0 ⟹ 2 x + y + 3 z + 7 = 0. 2(x-1)+1\!\left(y+\tfrac32\right)+3\!\left(z+\tfrac52\right)=0\;\Longrightarrow\;2x+y+3z+7=0. 2 ( x − 1 ) + 1 ( y + 2 3 ) + 3 ( z + 2 5 ) = 0 ⟹ 2 x + y + 3 z + 7 = 0. At C = ( 3 , 2 , − 5 ) C=(3,2,-5) C = ( 3 , 2 , − 5 ) 6 + 2 − 15 + 7 = 0 6+2-15+7=0 6 + 2 − 15 + 7 = 0
Step 4 — Radius.
r 2 = ∣ C A ∣ 2 = 38 ( 1 ) 2 − 32 ( 1 ) + 20 = 26 , r^{2}=|CA|^{2}=38(1)^{2}-32(1)+20=26, r 2 = ∣ C A ∣ 2 = 38 ( 1 ) 2 − 32 ( 1 ) + 20 = 26 , and as a check ∣ C B ∣ 2 = 38 − 18 + 6 = 26 |CB|^{2}=38-18+6=26 ∣ C B ∣ 2 = 38 − 18 + 6 = 26 r = 26 r=\sqrt{26} r = 26
Step 5 — Equation of the sphere.
( x − 3 ) 2 + ( y − 2 ) 2 + ( z + 5 ) 2 = 26. (x-3)^{2}+(y-2)^{2}+(z+5)^{2}=26. ( x − 3 ) 2 + ( y − 2 ) 2 + ( z + 5 ) 2 = 26. Expanding:
x 2 − 6 x + 9 + y 2 − 4 y + 4 + z 2 + 10 z + 25 = 26 , x^{2}-6x+9+y^{2}-4y+4+z^{2}+10z+25=26, x 2 − 6 x + 9 + y 2 − 4 y + 4 + z 2 + 10 z + 25 = 26 , x 2 + y 2 + z 2 − 6 x − 4 y + 10 z + 12 = 0. x^{2}+y^{2}+z^{2}-6x-4y+10z+12=0. x 2 + y 2 + z 2 − 6 x − 4 y + 10 z + 12 = 0. Step 6 — Verify both given points satisfy it.
At A = ( 0 , − 2 , − 4 ) A=(0,-2,-4) A = ( 0 , − 2 , − 4 ) 0 + 4 + 16 − 0 + 8 − 40 + 12 = 40 − 40 = 0 0+4+16-0+8-40+12=40-40=0 0 + 4 + 16 − 0 + 8 − 40 + 12 = 40 − 40 = 0
At B = ( 2 , − 1 , − 1 ) B=(2,-1,-1) B = ( 2 , − 1 , − 1 ) 4 + 1 + 1 − 12 + 4 − 10 + 12 = 0 4+1+1-12+4-10+12=0 4 + 1 + 1 − 12 + 4 − 10 + 12 = 0
Centre ( 3 , 2 , − 5 ) (3,2,-5) ( 3 , 2 , − 5 )
Answer
x 2 + y 2 + z 2 − 6 x − 4 y + 10 z + 12 = 0 , i.e. ( x − 3 ) 2 + ( y − 2 ) 2 + ( z + 5 ) 2 = 26 ; centre ( 3 , 2 , − 5 ) , r = 26 . \boxed{\;x^{2}+y^{2}+z^{2}-6x-4y+10z+12=0,\quad\text{i.e. }(x-3)^{2}+(y-2)^{2}+(z+5)^{2}=26;\ \ \text{centre }(3,2,-5),\ r=\sqrt{26}.\;} x 2 + y 2 + z 2 − 6 x − 4 y + 10 z + 12 = 0 , i.e. ( x − 3 ) 2 + ( y − 2 ) 2 + ( z + 5 ) 2 = 26 ; centre ( 3 , 2 , − 5 ) , r = 26 .