Question

The equation of the plane which cuts the sphere $x^2+y^2+z^2=a^2$ in a circle whose centre is $(\alpha ,\beta ,\gamma )$ is.

• A. $\alpha (x+\alpha )+\beta (y+\beta )+\gamma (z+\gamma )=0$
• B. $\alpha (x+\alpha )-\beta (y+\beta )-\gamma (z+\gamma )=0$
• C. $\alpha x+\beta y+\gamma z={\alpha }^2+{\beta }^2+{\gamma }^2$
• D. ${\alpha }^{2}x+{\beta }^{2}y+{\gamma }^{2}z=\alpha +\beta +\gamma$

Answer 1

The correct option is B $\alpha (x+\alpha )-\beta (y+\beta )-\gamma (z+\gamma )=0$

The equation of the sphere is

$x^2+y^2+z^2=a^2$

its centre $O(0,0,0)$ and radius= a

Centre of the circle is $M(\alpha ,\beta ,\gamma )$

Let the plane of the circle through $M(\alpha ,\beta ,\gamma )$be

$l(x-\alpha )+m(x-\beta )+n(x-\gamma )=0$

Since OM is perpendicular to plane

$\therefore$ normal is parallel to OM

$\frac{l}{\alpha }=\frac{m}{\beta }=\frac{n}{\gamma }$

Put these values of $l,m,n$, we get

$\therefore \alpha (x+\alpha )-\beta (y+\beta )-\gamma (z+\gamma )=0$