Respuesta :
Answer:
Equation of Sphere = [tex]x^{2}[/tex] + [tex]y^{2}[/tex] + [tex]z^{2}[/tex] - 18x - 4/3y +10z + 142/3 = 0
Step-by-step explanation:
Data Given:
P = (x,y,z)
Distance from P to A (-3,6,3) = Twice the distance from P to B(6,2,-3)
Solution:
Find the equation of the sphere:
It is given that:
PA = 2PB
Squaring both sides:
[tex](PA)^{2} = (2PB)^{2}[/tex]
[tex](PA)^{2} = 4 (PB)^{2}[/tex]
[tex](x - (-3))^{2}[/tex] + [tex](y - 6)^{2}[/tex] + [tex](z-3)^{2}[/tex] = 4 x [tex](x-6)^{2}[/tex] + [tex](y-2)^{2}[/tex] + [tex](z-(-3))^{2}[/tex]
Solving the above equation:
[tex](x + 3))^{2}[/tex] + [tex](y - 6)^{2}[/tex] + [tex](z-3)^{2}[/tex] = 4 x {[tex](x-6)^{2}[/tex] + [tex](y-2)^{2}[/tex] + [tex](z + 3))^{2}[/tex]}
[tex]x^{2}[/tex] + 9 + 6x + [tex]y^{2}[/tex] + 36 - 12y + [tex]z^{2}[/tex] + 9 - 6z = 4 { [tex]x^{2}[/tex] + 36 - 12x + [tex]y^{2}[/tex] + 4 - 4y + [tex]z^{2}[/tex] + 9 + 6z}
[tex]x^{2}[/tex] + 9 + 6x + [tex]y^{2}[/tex] + 36 - 12y + [tex]z^{2}[/tex] + 9 - 6z = 4[tex]x^{2}[/tex] + 144 - 48x +4[tex]y^{2}[/tex] + 16 - 16y + 4[tex]z^{2}[/tex] + 36 + 24z
Putting the right hand side = 0 and solving the equation:
[tex]x^{2}[/tex] - 4[tex]x^{2}[/tex] + [tex]y^{2}[/tex] - 4[tex]y^{2}[/tex] + [tex]z^{2}[/tex] - 4[tex]z^{2}[/tex] + 6x + 48x - 12y +16y -6z - 24z + 9 + 36 + 9 -144 - 16 - 36 = 0
-3[tex]x^{2}[/tex] - 3[tex]y^{2}[/tex] -3[tex]z^{2}[/tex] + 54x + 4y -30z -142 = 0
Taking (-) common
- ( 3[tex]x^{2}[/tex] + 3[tex]y^{2}[/tex] + 3[tex]z^{2}[/tex] - 54x - 4y +30z + 142) = 0
3[tex]x^{2}[/tex] + 3[tex]y^{2}[/tex] + 3[tex]z^{2}[/tex] - 54x - 4y +30z + 142 = 0
dividing the whole equation by 3
[tex]x^{2}[/tex] + [tex]y^{2}[/tex] + [tex]z^{2}[/tex] - 54/3x - 4/3y +30/3z + 142/3 = 0
[tex]x^{2}[/tex] + [tex]y^{2}[/tex] + [tex]z^{2}[/tex] - 54/3x - 4/3y +30/3z + 142/3 = 0
[tex]x^{2}[/tex] + [tex]y^{2}[/tex] + [tex]z^{2}[/tex] - 18x - 4/3y +10z + 142/3 = 0
The equation of the sphere for all surface point P such that [tex]AP = 2\cdot BP[/tex] is [tex]x^{2}+y^{2}+z^{2}-18\cdot x -\frac{4}{3}\cdot y + 10\cdot z +\frac{142}{3} = 0[/tex].
Procedure - Determination of the equation of a sphere
Geometric locus and Pythagorean expansion
Mathematically speaking, we have the following geometric locus:
[tex]AP = 2\cdot BP[/tex] (1)
Then, we expand the expression presented above by Pythagorean theorem, used to determine the length of a line segment in the field of analytical geometry:
[tex]\sqrt{(x +3)^{2} + (y-6)^{2} + (z-3)^{2}} = 2\cdot \sqrt{(x-6)^{2}+(y-2)^{2}+(z+3)^{2}}[/tex]
[tex](x+3)^{2}+(y-6)^{2}+(z-3)^{2} = 4\cdot (x-6)^{2}+4\cdot (y-2)^{2}+4\cdot (z+3)^{2}[/tex]
Resolution
Now we expand and simplify the resulting expression:
[tex]x^{2}+6\cdot x + 9 + y^{2}-12\cdot y + 36 + z^{2}-6\cdot z + 9 = 4\cdot (x^{2}-12\cdot x +36) + 4\cdot (y^{2}-4\cdot y +4) + 4\cdot (z^{2}+6\cdot z + 9)[/tex]
[tex]x^{2} + y^{2} + z^{2} + 6\cdot x -12\cdot y - 6\cdot z +54 = 4\cdot x^{2} +4\cdot y^{2} + 4\cdot z^{2} -48\cdot x -16\cdot y +24\cdot z +196[/tex]
[tex]3\cdot x^{2} + 3\cdot y^{2} + 3\cdot z^{2} -54\cdot x -4\cdot y + 30\cdot z + 142 = 0[/tex]
[tex]x^{2}+y^{2}+z^{2}-18\cdot x -\frac{4}{3}\cdot y + 10\cdot z +\frac{142}{3} = 0[/tex] (2)
This expression corresponds to the general form of the equation of the sphere for all surface point P such that [tex]AP = 2\cdot BP[/tex]. [tex]\blacksquare[/tex]
To learn more on geometric loci, we kindly invite to check this verified question: https://brainly.com/question/521553
