Question

Find an equation of the sphere with point P such that the distance from P to A(−2, 6, 2) is twice the distance from P to B(4, 3, −1).

Answer 1

Answer:

$3*x^{2} - 36*x + 3*y^{2} -12*y + 3*z^{2} -4*z + 62 = 0$

Step-by-step explanation:

The distance from a point $P = (x,y,z)$ to another point $A = (a,b,c)$ is given by $((x-a)^{2} + (y-b)^{2} + (z-c)^{2})^{1/2}$.

So, if the distance from P to A is twice the distance from P to B, we have an equality given by:

$((x+2)^{2} + (y-6)^{2} + (z-2)^{2})^{1/2} = 2*((x-4)^{2} + (y-3)^{2} + (z+1)^{2})^{1/2}$

If we power by 2 both sides of the equality to remove the square roots, we have:

$(x+2)^{2} + (y-6)^{2} + (z-2)^{2} = 4 * ((x-4)^{2} + (y-3)^{2} + (z+1)^{2})$

And solving every parenthesis and operating the similar terms, we have:

$(x^{2} + 4*x + 2) - 4*(x^{2} - 8*x + 16) + (y^{2} - 12 * y + 36) - 4*(y^{2} -6*y + 9) + (z^{2} -4*z+4) - 4*(z^{2}+2*z+1) = 0$

Operating and multiplying both sides by -1:

$(3*x^{2} - 36*x +62) + (3*y^{2} -12*y + 0) + (3*z^{2} -4*z + 0) = 0$

Giving as result:

$3*x^{2} - 36*x + 3*y^{2} -12*y + 3*z^{2} -4*z + 62 = 0$

The sphere can be seen here: https://www.wolframalpha.com/input/?i=3*x%5E2+-+36*x+%2B+3*y%5E2+-12*y+%2B+3*z%5E2+-4*z+%2B+62+%3D+0