If the centre of the sphere $$a{x^2} + b{y^2} + c{z^2} - 2x + 4y + 2z - 3 = 0$$ is $$\left( {\frac{1}{2},\, - 1,\, - \frac{1}{2}} \right),$$ what is the value of $$b\,?$$
A.
$$1$$
B.
$$ - 1$$
C.
$$2$$
D.
$$ - 2$$
Answer :
$$2$$
Solution :
The given equation of sphere is
$$a{x^2} + b{y^2} + c{z^2} - 2x + 4y + 2z - 3 = 0$$
This equation represents a equation of sphere, if coefficient of $${x^2},{y^2}$$ and $${z^2}$$ is same.
i.e., $$a = b = c$$
$$\therefore $$ Equation of sphere can be re-written as
$$\eqalign{
& b{x^2} + b{y^2} + b{z^2} - 2x + 4y + 2z - 3 = 0 \cr
& \Rightarrow {x^2} + {y^2} + {z^2} - \frac{{2x}}{b} + \frac{{4y}}{b} + \frac{{2z}}{b} - \frac{3}{b} = 0 \cr} $$
The centre of this sphere is $$\left( {\frac{1}{b},\,\frac{{ - 2}}{b},\,\frac{{ - 1}}{b}} \right)$$
Given that the centre of sphere is $$\left( {\frac{1}{2},\, - 1,\, - \frac{1}{2}} \right)$$
$$\frac{1}{b} = \frac{1}{2} \Rightarrow b = 2$$
Releted MCQ Question on Geometry >> Three Dimensional Geometry
Releted Question 1
The value of $$k$$ such that $$\frac{{x - 4}}{1} = \frac{{y - 2}}{1} = \frac{{z - k}}{2}$$ lies in the plane $$2x - 4y + z = 7,$$ is :
If the lines $$\frac{{x - 1}}{2} = \frac{{y + 1}}{3} = \frac{{z - 1}}{4}$$ and $$\frac{{x - 3}}{1} = \frac{{y - k}}{2} = \frac{z}{1}$$ intersect, then the value of $$k$$ is :
A plane which is perpendicular to two planes $$2x - 2y + z = 0$$ and $$x - y + 2z = 4,$$ passes through $$\left( {1,\, - 2,\,1} \right).$$ The distance of the plane from the point $$\left( {1,\,2,\,2} \right)$$ is :
Let $$P\left( {3,\,2,\,6} \right)$$ be a point in space and $$Q$$ be a point on the line $$\vec r = \left( {\hat i - \hat j + 2\hat k} \right) + \mu \left( { - 3\hat i + \hat j + 5\hat k} \right)$$
Then the value of $$\mu $$ for which the vector $$\overrightarrow {PQ} $$ is parallel to the plane $$x-4y+3z=1$$ is :