Question

If a sphere of constant radius k passes through $O(0,0,0)$ and meet the axes in $A,B,C$ and the centroid of $△ABC$ lies on $x^2+y^2+z^2=\lambda k^2$ then value of $\lambda$ equals

• A. $\frac{9}{4}$
• B. $\frac{4}{9}$
• C. $\frac{2}{3}$
• D. $\frac{3}{2}$

Answer 1

Answer: (B)

$\frac{4}{9}$

Let A,B,C are $(a,0,0),(0,b,0)$ and $(0,0,c)$ then equation of sphere OABC is

$|\overline{r}-(\frac{a}{2}\hat{i}+\frac{b}{2}\hat{j}+\frac{c}{2}\hat{k})|=\frac{1}{4}(a^2+b^2+c^2)$

$\therefore |\overline{r}-(\frac{a}{2}\hat{i}+\frac{b}{2}\hat{j}+\frac{c}{2}\hat{k})|={\left(\frac{1}{2}\sqrt{a^2+b^2+c^2}\right)}^2$

$\therefore k=Radiusofsphere=\frac{1}{2}\sqrt{a^2+b^2+c^2}$

$\therefore 4k^2=a^2+b^2+c^2$ ...$\left(1\right)$

Also centroid $(\alpha ,\beta ,\gamma )of△ABC$ gives

$\alpha =\frac{a+0+0}{3}$

$\therefore a=2\alpha ,b=3\beta ,c=3\gamma$

By $\left(1\right)$ we have $4k^2$

$=9({\alpha }^2+{\beta }^2+{\gamma }^2)$

$\Rightarrow (x^2+y^2+z^2)$

$=\frac{4}{9}{k}^2=\lambda k^2\left(Given\right)$

$\Rightarrow \lambda =\frac{4}{9}$