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2 years ago

7

Find an equation of the sphere containing all surface points P = (x, y, z) such that the distance from P to A(−3, 6, 3) is "twic

e the distance from P to" B(6, 2, −3).

1 answer:

Marina86 [1]2 years ago

7 0

Answer:

Equation of Sphere = $x^{2}$ + $y^{2}$ + $z^{2}$ - 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:

$(PA)^{2} = (2PB)^{2}$

$(PA)^{2} = 4 (PB)^{2}$

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

Solving the above equation:

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

$x^{2}$ + 9 + 6x + $y^{2}$ + 36 - 12y + $z^{2}$ + 9 - 6z = 4 { $x^{2}$ + 36 - 12x + $y^{2}$ + 4 - 4y + $z^{2}$ + 9 + 6z}

$x^{2}$ + 9 + 6x + $y^{2}$ + 36 - 12y + $z^{2}$ + 9 - 6z = 4 $x^{2}$ + 144 - 48x +4 $y^{2}$ + 16 - 16y + 4 $z^{2}$ + 36 + 24z

Putting the right hand side = 0 and solving the equation:

$x^{2}$ - 4 $x^{2}$ + $y^{2}$ - 4 $y^{2}$ + $z^{2}$ - 4 $z^{2}$ + 6x + 48x - 12y +16y -6z - 24z + 9 + 36 + 9 -144 - 16 - 36 = 0

-3 $x^{2}$ - 3 $y^{2}$ -3 $z^{2}$ + 54x + 4y -30z -142 = 0

Taking (-) common

- ( 3 $x^{2}$ + 3 $y^{2}$ + 3 $z^{2}$ - 54x - 4y +30z + 142) = 0

3 $x^{2}$ + 3 $y^{2}$ + 3 $z^{2}$ - 54x - 4y +30z + 142 = 0

dividing the whole equation by 3

$x^{2}$ + $y^{2}$ + $z^{2}$ - 54/3x - 4/3y +30/3z + 142/3 = 0

$x^{2}$ + $y^{2}$ + $z^{2}$ - 54/3x - 4/3y +30/3z + 142/3 = 0

$x^{2}$ + $y^{2}$ + $z^{2}$ - 18x - 4/3y +10z + 142/3 = 0

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