UPSC 2024 Maths Optional Paper 1 Q1e — Step-by-Step Solution

10 marks · Section A

Cylinder · Analytic Geometry · asked 4× in 13 yrs · Read the full method →

Question

Find the equation of the right circular cylinder which passes through the circle $x^2+y^2+z^2=9$ , $x-y+z=3$ .

Technique

The cylinder’s axis runs along the plane’s normal through the circle’s centre; a point lies on the cylinder iff its squared perpendicular distance from the axis equals the circle’s radius squared.

Solution

$Sphere x^2+y^2+z^2=9 (centre O) cut by the plane x-y+z=3 in a circle of radius \sqrt6 centred at C=(1,-1,1), where OC=\sqrt3. The required right circular cylinder has its axis along the plane's normal (1,-1,1)/\sqrt3 through C, and the same radius \sqrt6 — every point on it lies at perpendicular distance \sqrt6 from that axis.$

Step 1 — Locate the circle.

The circle is the intersection of the sphere $x^2+y^2+z^2=9$ (centre $O=(0,0,0)$ , radius $3$ ) with the plane $x-y+z=3$ .

Distance from $O$ to the plane: $d=|0-0+0-3|/\sqrt{1+1+1}=3/\sqrt{3}=\sqrt{3}$ .

Circle radius: $\rho=\sqrt{9-3}=\sqrt{6}$ .

Centre of circle (foot of perpendicular from $O$ to the plane): along the unit normal $\hat n=(1,-1,1)/\sqrt{3}$ at distance $\sqrt{3}$ , giving $C=\sqrt{3}\cdot\hat n=(1,-1,1)$ .

Step 2 — Set up the cylinder condition.

A point $P=(x,y,z)$ lies on the cylinder iff its perpendicular distance from the axis (line through $C$ along $\hat n$ ) equals $\rho=\sqrt{6}$ . This gives

|\vec{CP}|^2-\left(\frac{\vec{CP}\cdot\hat n}{|\hat n|}\right)^2=6.

Here $\vec{CP}=(x-1,y+1,z-1)$ and $\vec{CP}\cdot\hat n/|\hat n|=(x-y+z-3)/\sqrt{3}$ , so

(x-1)^2+(y+1)^2+(z-1)^2-\frac{(x-y+z-3)^2}{3}=6.

Step 3 — Expand and simplify.

$(x-1)^2+(y+1)^2+(z-1)^2=x^2+y^2+z^2-2x+2y-2z+3.$

$(x-y+z-3)^2=(x-y+z)^2-6(x-y+z)+9.$

Multiplying the equation through by $3$ :

3(x^2+y^2+z^2-2x+2y-2z+3)-(x-y+z)^2+6(x-y+z)-9=18.

Expanding $(x-y+z)^2=x^2+y^2+z^2-2xy+2xz-2yz$ :

3(x^2+y^2+z^2)-6x+6y-6z+9-(x^2+y^2+z^2-2xy+2xz-2yz)+6x-6y+6z-9=18.

2(x^2+y^2+z^2)+2xy-2xz+2yz=18.

Divide by $2$ :

Answer

\boxed{\;x^2+y^2+z^2+xy-xz+yz=9.\;}