The plane $x+2y-z=4$ cuts the sphere $x^2+y^2+z^2-x+z-2=0$ in a circle of radius?
I tried putting value of y from plane in sphere but then I get a $zx$ term. How to proceed?
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$x^2+y^2-x=(z+2)(1-z)$ – Aditya Agarwal Dec 12 '15 at 13:14
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It doesn't cut it in a circle, its an ellipse. – Aditya Agarwal Dec 12 '15 at 13:16
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@AdityaAgarwal How to prove that it is an ellipse? – Mathematics Dec 12 '15 at 13:16
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See my answer below, simplify the equation. – Aditya Agarwal Dec 12 '15 at 13:20
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@Mirko Question is looking for the radius of intersection circle – Mathematics Dec 12 '15 at 14:00
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one idea, to simplify things, once you know the radius of the sphere and the distance from its center to the center of the intersecting circle, translate/rotate so that the sphere is centered at the origin, and the intersection circle is centered on the $z$ axis, the equations will be simple and easy to solve. That will give you the radius, without giving you the equation of the actual intersecting circle. – Mirko Dec 12 '15 at 14:04
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First a normal vector to the plane is $(1,2,-1)$. The sphere equation could be rewritten as $\left(x-\dfrac12\right)^2+y^2+\left(z+\dfrac12\right)^2=\dfrac52$, hence the center is $\left(\dfrac12,0,\dfrac{-1}2\right)$. The center of the intersection circle is on the line $\left(\dfrac12+t,2t,\dfrac{-1}2-t\right)$, satisfying $x+2y-z=4$, that is $\dfrac12+t+4t+\dfrac12+t=4$ from which you get $6t=3$, $t=\dfrac12$. Thus the center of the intersection circle is $(1,1,-1)$ and from here with a little(hmm?) more effort you could find the equation of the circle.
Perhaps easier (given you are only looking for the radius, but not for the equation of the intersection circle), do not try to find the actual equation of the intersection circle (which might be tedious) but replace the sphere with one through the origin, and replace the plane with a horizontal plane (i.e. perpendicular to the $z$ axis). As long as the sphere has the same radius, and the center of the intersecting circle is the same distance from the center of the sphere, then the radius of the intersection circle would be the same.
Note that the distance of the two centers in question is $\sqrt{\left(\dfrac12\right)^2+1^2+\left(\dfrac12\right)^2}=\sqrt{\dfrac32}$. Consider the new sphere $x^2+y^2+z^2=\dfrac52$ and new plane $z=\sqrt{\dfrac32}$. Substituting $z$ from the equation of the (new) plane into the (new) sphere gives you the equation $x^2+y^2=1$, which radius is, of course, $1$.
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As one answer already pointed out, first complete the squares to simplify the equation of the sphere: $$(x-0.5)^2+y^2+(z+0.5)^2=2.5$$
So the centre of the sphere is at $(0.5,0,-0.5)$ and its radius is $R=\sqrt{\dfrac 52}$
Now say the distance of the plane from the centre of sphere is $d$. Simply applying the necessary formula we get $d=\sqrt{\dfrac 32}$
Now just draw a diagram of the situation and you'll find that the radius of the circle of intersection $r$ is simply $$r=\sqrt{R^2-d^2}=1$$
najayaz
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1@Mathematics yes it will. See Andrew's answer on this post. I think a lot can be learned from it. – najayaz Dec 12 '15 at 14:33
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$x+2y-4=z$ and $x^2+y^2-x=-(z^2+z-2)$ $$x^2+y^2-x=(z+2)(1-z)$$ Substituting, $$x^2+y^2-x=(x+2y-2)(1-(x+2y-4))$$ $$x^2+y^2-x=-(x+2y-2)(x+2y-5)$$ And this way you get an equation in $x(s)$ and $y$(s).
Aditya Agarwal
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The intersection of a plane and a sphere in 3-D space is ALWAYS a circle. By eliminating $z$ from the equations you have just obtained the projection of that circle on the XY-plane, which is obviously an ellipse. – najayaz Dec 12 '15 at 14:25