← 2023 Paper 1

UPSC 2023 Maths Optional Paper 1 Q3c — Step-by-Step Solution

15 marks · Section A

Sphere · Analytic Geometry · asked 17× in 13 yrs · Read the full method →

Question

Find the equation of the sphere through the circle x2+y2+z2−4x−6y+2z−16=0;  3x+y+3z−4=0x^2+y^2+z^2-4x-6y+2z-16=0;\;3x+y+3z-4=0 in the following two cases.

(i) the point (1,0,−3)(1,0,-3) lies on the sphere.

(ii) the given circle is a great circle of the sphere.

Technique

Family of spheres through a circle is S+λL=0S+\lambda L=0. (i) substitute the given point to find λ\lambda. (ii) “great circle” ⇔\Leftrightarrow plane passes through centre; use the centre formula and substitute.

Solution

Family of spheres through the circle:

S+λL=0,S+\lambda L=0,

where S≡x2+y2+z2−4x−6y+2z−16S\equiv x^2+y^2+z^2-4x-6y+2z-16 is the original sphere and L≡3x+y+3z−4L\equiv 3x+y+3z-4 is the plane.

Every sphere in this family passes through the given circle.

Case (i) — passing through (1,0,−3)(1,0,-3)

Plug into S+λL=0S+\lambda L=0:

1+0+9−4−0−6−16+λ(3+0−9−4)=0.1+0+9-4-0-6-16+\lambda(3+0-9-4)=0. −16+λ(−10)=0⇒λ=−1610=−85.-16+\lambda(-10)=0\Rightarrow\lambda=-\tfrac{16}{10}=-\tfrac{8}{5}.

Sphere equation:

x2+y2+z2−4x−6y+2z−16−85(3x+y+3z−4)=0.x^2+y^2+z^2-4x-6y+2z-16-\tfrac{8}{5}(3x+y+3z-4)=0.

Multiply by 55:

  5(x2+y2+z2)−44x−38y−14z−48=0.  \boxed{\;5(x^2+y^2+z^2)-44x-38y-14z-48=0.\;}

(Verification: at (1,0,−3)(1,0,-3): 5(1+0+9)−44−0+42−48=50−44+42−48=05(1+0+9)-44-0+42-48=50-44+42-48=0 ✓.)

Case (ii) — given circle is a great circle of the sphere

A great circle is one whose plane passes through the centre of the sphere. So the centre of the sphere lies on the plane 3x+y+3z=43x+y+3z=4.

The centre of S+λL=x2+y2+z2+(−4+3λ)x+(−6+λ)y+(2+3λ)z+(−16−4λ)=0S+\lambda L=x^2+y^2+z^2+(-4+3\lambda)x+(-6+\lambda)y+(2+3\lambda)z+(-16-4\lambda)=0 is

C=(4−3λ2, 6−λ2, −2+3λ2).C=\Bigl(\tfrac{4-3\lambda}{2},\,\tfrac{6-\lambda}{2},\,-\tfrac{2+3\lambda}{2}\Bigr).

Plug into 3x+y+3z=43x+y+3z=4:

3⋅4−3λ2+6−λ2+3⋅(−2+3λ2)=4.3\cdot\tfrac{4-3\lambda}{2}+\tfrac{6-\lambda}{2}+3\cdot\bigl(-\tfrac{2+3\lambda}{2}\bigr)=4. 12−9λ+6−λ−6−9λ2=4.\tfrac{12-9\lambda+6-\lambda-6-9\lambda}{2}=4. 12−19λ2=4⇒12−19λ=8⇒λ=419.\tfrac{12-19\lambda}{2}=4\Rightarrow 12-19\lambda=8\Rightarrow \lambda=\tfrac{4}{19}.

Sphere equation:

x2+y2+z2−4x−6y+2z−16+419(3x+y+3z−4)=0.x^2+y^2+z^2-4x-6y+2z-16+\tfrac{4}{19}(3x+y+3z-4)=0.

Multiply by 1919:

Answer

  19(x2+y2+z2)−64x−110y+50z−320=0.  \boxed{\;19(x^2+y^2+z^2)-64x-110y+50z-320=0.\;}
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