Question

A sphere of constant radius 'K' passes through origin and meets axes in A,B,C. The centroid of the $△ABC$ lies on the sphere

• A. $9(x^2+y^2+z^2)=4K^2$
• B. $3(x^2+y^2+z^2)=4K^2$
• C. $(x^2+y^2+z^2)=4K^2$
• D. $x^2+y^2-z^2=9K^2$

Answer 1

The correct option is A $9(x^2+y^2+z^2)=4K^2$

Let the equation of the sphere be

$(x-\alpha {)}^2+(y-\beta {)}^2+(z-\gamma {)}^2=k^2$

Since it passes through the origin

$\therefore {\alpha }^2+{\beta }^2+{\gamma }^2=k^2$ and meets the coordinate axis

at the point $A(2\alpha ,0,0),\beta (0,2\beta ,0),c(0,0,2\gamma )$

The centroid of the triangle ABC is

$(\frac{2\alpha }{2},\frac{2\beta }{3},\frac{2\gamma }{3})=(u,v,w)$

Then $u^2+v^2+w^2=\left(\frac{4}{9}\right)({\alpha }^2+{\beta }^2+{\gamma }^2)=\left(\frac{4}{9}\right)=k^2$

Thus $(u,v,w)$ lies on $9(x^2+y^2+z^2)=4k^2$