Find the intersection of the spheres x2y2z20 adn x42y22z429S

Find the intersection of the spheres x²+y²+z²=0 adn (x-4)²+(y+2)²+(z-4)²=9

Solution

Equation of Spehre1 : x² + y² + z² = 0

Since , we know that the square of any number is greater than or equal to zero and can be equal to zero if and only if the number is equal to zero , it implies that

x = 0 , y = 0 and z = 0

Equation of Spehre2 : (x - 4)² + (y + 2)² + (z - 4)² = 3²

Center of Sphere2 : < 4 , -2 , 4 >

Radius of Sphere2 : 3

Distance between Center of Sphere2 and Center of Sphere1 => ( ( 4 - 0 )² + ( -2 - 0 )² + ( 4 - 0 )² ) = 6

Since distance between the centers of both spheres is greater than sum of the radius of both spheres , hence both spheres do not intersect