Question

Equation of the sphere with center in the positive octant which passes through the circle ${x^2} + {y^2} = 4$,$z = 0$ and is cut by the plane $x + 2y + 2z = 0$ in a circle of radius $3$ is
A.\[{x^2} + {y^2} + {z^2} - 6x - 4 = 0\]
B.\[{x^2} + {y^2} + {z^2} - 6z + 4 = 0\]
C.\[{x^2} + {y^2} + {z^2} - 6z - 4 = 0\]
D.\[{x^2} + {y^2} + {z^2} - 6y - 4 = 0\]

Answer 1

Hint: A geometrical object in a three-dimensional space which resembles the surface of a ball is called a sphere. It is defined as the set of all those points which are at a same distance from a given point. This given point is referred to as the center of the sphere. The distance between the center and any point on the surface of the sphere is known as radius, which is represented by the letter ‘r’.

Complete step by step answer:
We know that the distance between the center and any point on the surface of the sphere is called radius i.e., ‘r’ and double the radius is called diameter.
The diameter divides the sphere into two equal parts, which are known as hemispheres. The general equation of a sphere is \[{x^2} + {y^2} + {z^2} = {r^2}\].
Now, according to the question,
The equation of the sphere through the given circle is given by \[{x^2} + {y^2} + {z^2} - 4 + \lambda z = 0\].
The center of this sphere is \[\left( {0,0, - \dfrac{\lambda }{2}} \right)\] whereas
The radius is given by \[r = \sqrt {0 + 0 + \dfrac{{{\lambda ^2}}}{4} + 4} \].
Let us assume that the ‘d’ is the distance of the plane \[x + 2y + 2z = 0\] from the center of the sphere.
So,\[d = \dfrac{{|0 + 2 \times 0 + 2\left( { - \dfrac{\lambda }{2}} \right)|}}{{\sqrt {1 + {2^2} + {2^2}} }}\]
\[d = \dfrac{{|\lambda |}}{3}\]
It is given that the sphere cuts the plane in a circle whose radius is of \[3\]. So,
\[
   \Rightarrow {r^2} - {d^2} = {3^2} \\
   \Rightarrow \dfrac{{{\lambda ^2}}}{4} + 4 - \dfrac{{{\lambda ^2}}}{9} = 9 \\
   \Rightarrow \dfrac{{5{\lambda ^2}}}{{36}} = 5 \\
 \]
\[
   \Rightarrow {\lambda ^2} = 36 \\
   \Rightarrow \lambda = \pm 6 \\
 \]
We are given that the center lies in the positive octant, which means that \[\lambda < 0\]
Therefore, we can conclude that the required equation is \[{x^2} + {y^2} + {z^2} - 6z - 4 = 0\].

Hence, option (C) is correct.

Note:
For deriving the equation of a sphere from a circle whose origin is at \[\left( {0,0} \right)\], the formula is given by \[{x^2} + {y^2} = {r^2}\]. In order to solve such types of questions students should keep the formulas in their mind. The volume of a sphere can be calculated using formula $V = \dfrac{4}{3}\pi {r^3}$ whereas the surface area can be calculated using $A = 4\pi {r^2}$.