# Thread: Intersection of two sphere

1. ## 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?

2. 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.

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

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

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 $\sqrt{3}\leq |x| \leq \sqrt{15}$.

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

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