# Intersection of two sphere

• Nov 4th 2010, 08:06 PM
hashshashin715
Intersection of two sphere
Fid the constant c such that at any point of intersection of the two shpere

(x - c)^2 + y^2 + z^2 = 3 and x^2 + (y - 1)^2 + z^2 = 1

the corresponding tangent plane will be perpendicular to each other.

Info for you: the the gradient of f is normal to the tangent plane of a level surface f(x) = c.

So if the two gradient vectors are perpendicular, then the planes should be too. I found the two gradient vectors and dot product(ed) them together and set them to 0. But I don't know what the intersection is for (I'm getting an equation of a line for the intersection which doesn't make sense). Any help?
• Nov 5th 2010, 01:29 AM
earboth
Quote:

Originally Posted by hashshashin715
Fid the constant c such that at any point of intersection of the two shpere

(x - c)^2 + y^2 + z^2 = 3 and x^2 + (y - 1)^2 + z^2 = 1

the corresponding tangent plane will be perpendicular to each other.

...

1. The sphere $\displaystyle s_1: (x - c)^2 + y^2 + z^2 = 3$ has it's center at $\displaystyle C_1(c,0,0)$ and the constant radius $\displaystyle r_1=3$.

The sphere $\displaystyle s_2: x^2 + (y-1)^2 + z^2 = 1$ has it's center at $\displaystyle C_2(0,1,0)$ and the constant radius $\displaystyle r_2=1$.

2. The two centers are located in the x-y-plane. I've made a sketch of the situation. (see attachment)

3. To get at least points of intersections the two centers must have a distance between 4 and 2. Use Pythagorean theorem to determine the value of c.
Ive got $\displaystyle \sqrt{3}\leq |x| \leq \sqrt{15}$.

4. If $\displaystyle |c|=\sqrt{3}$ the smaller sphere touches the larger sphere from the interior; if $\displaystyle |c|=\sqrt{15}$ the smaller sphere touches the larger sphere from the exterior.

5. If $\displaystyle \sqrt{3}< |x| < \sqrt{15}$ then the points of intersection form a circle.