From the point (1,-2,3) lines are drawn to meet the sphere $x^{2} + y^{2} + z^{2} = 4$ and they are divided internally in the ratio 2:3. The locus of the point of division is-

$5 x^{2} + 5 y^{2} + 5 z^{2} - 6 x + 12 y + 22 = 0$

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$5 (x^{2} + y^{2} + z^{2}) = 22$

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$5 x^{2} + 5 y^{2} + 5 z^{2} - 2 x y - 3 y z z x - 6 x + 12 y + 5 z + 22 = 0$

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$5 x^{2} + 5 y^{2} + 5 z^{2} - 6 x + 12 y - 18 z + 22 = 0$

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Solution

The correct option is D
$5 x^{2} + 5 y^{2} + 5 z^{2} - 6 x + 12 y - 18 z + 22 = 0$

Suppose any line through the given point (1,-2,3) meet the sphere $x^{2} + y^{2} + z^{2} = 4$ in the point $(x_{1}, y_{1}, z_{1}) then x_{1}^{2} + y_{1}^{2} + z_{1}^{2} = 4$
Now let the coordinate of the point which divides the join of (1,-2,3) and $(x_{1}, y_{1} z_{1}) in the ratio 2:3 be x_{2}, y_{2}, z_{2} then we have \begin{matrix} x_{2} = & \frac{2 x_{1} + 3. (1)}{2 + 3} \Rightarrow & x_{1} = \frac{5 x_{2} - 3}{2} y_{2} = & \frac{2. y_{1} + 3 (- 2)}{2 + 3} \Rightarrow & y_{1} = \frac{5 y_{2} + 6}{2} z_{2} = & \frac{2. z_{1} + 3.3}{2 + 3} \Rightarrow & z_{1} = \frac{5 z_{2} - 9}{2} \end{matrix} ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭$
Putting value of $x_{1}, y_{1}, z_{1} from (2) in (1) , we have (5 x_{2} - 3)^{2} + (5 y_{2} + 6)^{2} + (5 z_{2} - 9)^{2} = 4 \times 4 \Rightarrow 25 (x_{2}^{2} + y_{2}^{2} + z_{2}^{2}) - 30 x_{2} + 60 y_{2} \div 90 z_{2} + 110 = 0 \Rightarrow 5 (x_{2}^{2} + y_{2}^{2} + z_{2}^{2}) - 6 (x_{2} - 2 y_{2} + 3 z_{2}) + 22 = 0$

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Similar questions

(1, - 2, 3),

x^{2} + y^{2} + z^{2} = 4

2 : 3

x^{2} + y^{2} = a^{2}

A (- 1, 1)

P

x^{2} + y^{2} = 4

3 : 2

y = x

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

1 : 2

\frac{x^{2}}{4} - \frac{y^{2}}{9} = 1

x^{2} + y^{2} = 4

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