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A sphere of constant radius $\mathrm{r}$ through the origin intersects the coordinate axes in $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$. What is the locus of the centroid of the triangle ABC?
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The correct answer is:
$9\left(x^{2}+y^{2}+z^{2}\right)=4 r^{2}$
Let the sphere passing through points $\mathrm{A}(\mathrm{a}, 0,0), \mathrm{B}(0,$, $\mathrm{b}, 0), \mathrm{C}(0,0, \mathrm{c})$
Equation of sphere is $\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}-\mathrm{ax}-\mathrm{by}-\mathrm{cz}=0$
radius, $\mathrm{r}=\frac{1}{2} \sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}}$
$\Rightarrow a^{2}+b^{2}+c^{2}=4 r^{2}$
(Squaring on both sides)
Let $(\alpha, \beta, \gamma)$ be centroid of sphere. $(\alpha, \beta, \gamma)=\left(\frac{a+0+0}{3}, \frac{0+b+0}{3}, \frac{0+0+c}{3}\right)$
$=\left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right)$
$\alpha^{2}+\beta^{2}+\gamma^{2}=\frac{\mathrm{a}^{2}}{9}+\frac{\mathrm{b}^{2}}{9}+\frac{\mathrm{c}^{2}}{9}$
$=\frac{a^{2}+b^{2}+c^{2}}{9}$
$($ from $(1))$
$\left.z^{2}\right)=4 r^{2}$
Equation of sphere is $\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}-\mathrm{ax}-\mathrm{by}-\mathrm{cz}=0$
radius, $\mathrm{r}=\frac{1}{2} \sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}}$
$\Rightarrow a^{2}+b^{2}+c^{2}=4 r^{2}$
(Squaring on both sides)
Let $(\alpha, \beta, \gamma)$ be centroid of sphere. $(\alpha, \beta, \gamma)=\left(\frac{a+0+0}{3}, \frac{0+b+0}{3}, \frac{0+0+c}{3}\right)$
$=\left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right)$
$\alpha^{2}+\beta^{2}+\gamma^{2}=\frac{\mathrm{a}^{2}}{9}+\frac{\mathrm{b}^{2}}{9}+\frac{\mathrm{c}^{2}}{9}$
$=\frac{a^{2}+b^{2}+c^{2}}{9}$
$($ from $(1))$
$\left.z^{2}\right)=4 r^{2}$
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