If $(2,3,-3)$ is one end of a diameter of the sphere $x^2+y^2+z^2-6 x-12 y-2 z+20=0$, then the other end of the diameter is

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Aicharya · Accepted Answer

The equation of the sphere is $x^2+y^2+z^2-6 x-12 y-2 z+20=0$

So, the centre of the sphere is $(-u,-v,-w)=(3,6,1)$ given the one end of the diameter is $(2,3,-3)$, let the other end of the diameter is $(\alpha, \beta, \gamma)$. Since, $O$ is the mid point of the diameter, then $(3,6,1)=\left[\frac{\alpha+2}{2}, \frac{\beta+3}{2}, \frac{\gamma-3}{2}\right]$ $\Rightarrow \quad \frac{\alpha+2}{2}=3, \alpha=4$ $\Rightarrow \quad \frac{\beta+3}{2}=6, \beta=9$ $\Rightarrow \quad \frac{\gamma-3}{2}=1, \gamma=5$ Hence, the other end is $(4,9,5)$.

If $(2,3,-3)$ is one end of a diameter of the sphere $x^2+y^2+z^2-6 x-12 y-2 z+20=0$, then the other end of…

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