0

An equation of the sphere is given:$$x^2+y^2+z^2+2x-4y-2z+1=0$$ then the centre of the sphere lies on the plane $2x-4y-2z+1=0$ or not ? My attempt - the the centre of the sphere is $(-1,2,1)$ which is not equal to DRS of the plane.So the centre of the sphere doesn't lies on the plane.Am correct ?

zoli
  • 20,452
mSourav
  • 183
  • 1
    Yep.... That's right! – David G. Stork May 23 '18 at 14:46
  • @mSourav Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work – user Jun 22 '18 at 20:13

2 Answers2

1

The sphere $$x^2+y^2+z^2+2x-4y-2z+1=0 \implies$$ $$(x+1)^2+(y-2)^2+(z-1)^2=5$$

and the plane

$$2x-4y-2z+1=0 \implies $$ $$(-1)x+(2)y+(1)z=\frac{1}{2}$$

since $\frac{1}{2} \neq (-1)^2+(2)^2+(1)^2$ the plane doesn't pass through the center of the circle. However the plane is perpendicular to the vector from the origin to the center of the circle.... for whatever that's worth....

MaximusFastidiousIrreverence
  • 1,735
0

We have that the system for the intersection sphere-plane

  • $x^2+y^2+z^2+2x-4y-2z+1=0$
  • $2x-4y-2z+1=0$

implies that

  • $x^2+y^2+z^2=0 \implies x=y=z=0$

which is a single point, notably the origin $\not \in$ sphere and plane, then the center can't be on that plane.

user
  • 154,566