Question

Equation of the sphere with centre in the positive octant which passes through the circle x² + y² = 4, z = 0 and is cut by the plane x + 2y + 2z = 0 in a circle of radius 3 is –

• A. x² + y² + z² – 6x – 4 = 0
• B. x² + y² – 6y – 4 = 0
• C. x² + y² + z² – 6z – 4 = 0
• D. x² + y² – 6x – 6y – 4 = 0

Answer 1

Answer: (C)

x² + y² + z² – 6z – 4 = 0

Answer 2

Equation of sphere through the given circle is

x² + y² + z² – 4 + lz = 0 … (1)

Its centre is $\left( 0,0,–\frac{\lambda}{2} \right)$ and radius $\sqrt{0 + 0 + \frac{\lambda^{4}}{4} + 4}$

\ d = distance of the plane x +2y + 2z = 0 from the centre of

sphere

= $\frac{\left| 0 + 2 \times 0 + 2–\left( \frac{\lambda}{2} \right) \right|}{\sqrt{1 + 4 + 4}}$ = $\frac{\lambda}{3}$

\ Since the sphere (1) cuts the plane in a circle of radius 3

\ r² – d² = 3² Ž $\frac{\lambda^{2}}{4}$+ 4 – $\frac{\lambda^{2}}{9}$= 9

Ž l ± 6

Since the centre lies in the positive octant, so l < 0

\ Required equation is x² + y² + z² – 6z + 4 = 0.