Question

Coordinates of centre of sphere $6x^2+6y^2+6z^2-16x+9z-6=0$ are.

• A. $\left(\frac{3}{4},0,\frac{4}{3}\right)$
• B. $(-16,0,9)$
• C. $\left(-8,0,\frac{9}{2}\right)$
• D. $\left(\frac{4}{3},0,\frac{-3}{4}\right)$

Answer 1

The correct option is D $\left(\frac{4}{3},0,\frac{-3}{4}\right)$

For the sphere, the standard equation can be written as $(x-a{)}^2+(y-b{)}^2+(z-c{)}^2=r^2$

For this sphere, the center is $(a,b,c)$

When expanded, we have the equation as $x^2-2ax+y^2-2by+z^2-2cz+a^2+b^2+c^2-r^2=0$

The given equation reads $6x^2+6y^2+6z^2-16x+9z-6=0$

When divided throughout by $6$, we get $x^2+y^2+z^2-\frac{16}{6}x+\frac{9}{6}z-1=0$

The center coordinates thus become $(\frac{16}{12},0,\frac{-9}{12})$ i.e. $(\frac{4}{3},0,\frac{-3}{4})$