The point at which the axes are translated to remove first degree terms is
List - I
List - II

$A) x^{2} + y^{2} + z^{2} - 2 x$
$- 4 y + 2_{Z} - 3 = 0$

1. $(- 1, - 2, 1)$

$B) x^{2} + y^{2} - z^{2} + 2 x$
$+ 4 y + 2_{Z} + 3 = 0$

2. $($ 1, 2, 1 $)$

$C) 2 x^{2} - 2 y^{2} + z^{2} - 4 x$
$+ 8 y - 2_{Z} - 5 = 0$

$3. (- 1, 1, 2)$

$D) 2 x^{2} - 3 y^{2} + z^{2} + 4 x$
$+ 6 y - 4_{Z} - 2 = 0$

$4. (1, 2, - 1)$

4 1 2 3

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1 2 3 4

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Solution

The correct option is C 4 1 2 3
A) $x^{2} + y^{2} + z^{2}$ -2x-4y+2z-3 $=$ 0
${(x - 1)}^{2} + {(y - 2)}^{2} + {(z + 1)}^{2}$ -9 $=$ 0
So axis is translated to (1,2,-1)
B)
$x^{2} + y^{2} - z^{2}$ +2x+4y+2z+3 $=$ 0
${(x + 1)}^{2} + {(y + 2)}^{2} - {(z - 1)}^{2}$ -1 $=$ 0
So axis is translated to (-1,-2,1)
C)
$2 x^{2} - 2 y^{2} + z^{2}$ -4x+8y-2z-5 $=$ 0
$2 {(x - 1)}^{2} - 2 {(y - 2)}^{2} + {(z - 1)}^{2}$ $=$ 0
So axis is translated to (1,2,1)
D)
$2 x^{2} - 3 y^{2} + z^{2}$ +4x+6y-4z-2 $=$ 0
$2 {(x + 1)}^{2} - 3 {(y - 1)}^{2} + {(z - 2)}^{2}$ -5 $=$ 0
So axis is translated to (-1,1,2)
therefore option A is correct

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