Can a circle have a negative radius?

Question

I am working on a problem that asks if a given sphere intersects the zx-plane.

The equation of the sphere is $(x-2)^2+(y+6)^2+(z-4)^2=5^2$

Can someone please explain to me how the sphere does not intersect the zx-plane because the radius of the circle is said to be a positive quantity when in fact there is clearly a negative sign in front of the 11.

Thank you

Answer 1

This is called a proof by contradiction.

You first do the assumption that your sphere intersects the plane. From this you would deduce that $(x-2)^2 + (z-4)^2 = -11$. But this is not possible since the number on the left is non-negative and the one on the right is negative. Hence you deduce that your initial assumption was false: the sphere does not intersect the plane.

Answer 2

For any $a\in\Bbb R$ we have $a^{2}\geq0$ and it follows that $(x-2)^{2}+(z-4)^{2}\geq0$. However since you've shown that $(x-2)^{2}+(z-4)^{2}=-11$ we have a contradiction. Clearly the radius should be a positive quantity as mentioned by Blue, so the sphere does not intersect the $zx$-plane.

Answer 3

If your sphere intersect the $xz$-plane, then there is a point $[x, y, z]$, which is both a point of the $xz$-plane (i. e. $y = 0$) and a point of your sphere (i. e. it satisfied the equation of the sphere).

But then, consequently, $(x-2)^2 + (z-4)^2$ must be a negative number, which is impossible.

So the assumption that the sphere intersect the $xz$-plane was wrong.

Answer 4

Use the equations to determine their radius and centers. The radius of sphere is 5, it's center is at $x=2 , y=-6 , z=4$ The distance from xz plane is to center of sphere is 6, the radius of sphere is 5 , it is 1 unit short of touching the xz plane.