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If the centre of the sphere $a x^{2}+b y^{2}+c z^{2}-2 x+4 y+2 z-3=0$ is $(1 / 2,-1,-1 / 2)$, what
is the value of $\mathrm{b}$ ?
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is the value of $\mathrm{b}$ ?
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Verified Answer
The correct answer is:
2
The given equation of sphere $a x^{2}+b y^{2}+c z^{2}-2 x+4 y+2 z-3=0$
This equation represents a equation of sphere, if coefficient of $x^{2}, y^{2}$ and $z^{2}$ is same.
i.e., $a=b=c$
Equation of sphere can be re-written as $b x^{2}+b y^{2}+b z^{2}-2 x+4 y+2 z-3=0$
$\Rightarrow x^{2}+y^{2}+z^{2}-\frac{2 x}{b}+\frac{4 y}{b}+\frac{2 z}{b}-\frac{3}{b}=0$
The centre of this sphere is $\left(\frac{1}{b}, \frac{-2}{b}, \frac{-1}{b}\right)$
Given that the centre of sphere is $\left(\frac{1}{2},-1,-\frac{1}{2}\right)$
$\frac{1}{b}=\frac{1}{2} \Rightarrow b=2$
This equation represents a equation of sphere, if coefficient of $x^{2}, y^{2}$ and $z^{2}$ is same.
i.e., $a=b=c$
Equation of sphere can be re-written as $b x^{2}+b y^{2}+b z^{2}-2 x+4 y+2 z-3=0$
$\Rightarrow x^{2}+y^{2}+z^{2}-\frac{2 x}{b}+\frac{4 y}{b}+\frac{2 z}{b}-\frac{3}{b}=0$
The centre of this sphere is $\left(\frac{1}{b}, \frac{-2}{b}, \frac{-1}{b}\right)$
Given that the centre of sphere is $\left(\frac{1}{2},-1,-\frac{1}{2}\right)$
$\frac{1}{b}=\frac{1}{2} \Rightarrow b=2$
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