Question

The equation of a sphere is $3x^{2}+3y^{2}+3z^{2}+12x+18y+24z+30=0.$ The centre and radius of sphere, respectively, are:

• A. (3, 2, 4) and $\sqrt{29}$
• B. (3, -2, -4) and $\sqrt{19}$
• C. (-3, 2, -4) and $\sqrt{29}$
• D. $(-2, -3, -4)$ and $\sqrt{19}$

Hint:

Always ensure the coefficients of $x^2, y^2,$ and $z^2$ are equal to 1 before identifying $u, v,$ and $w$. If they are not 1, divide the entire equation by that coefficient first.

Answer 1

The Correct Option is D

Concept: The standard general equation of a sphere is $x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d = 0$. For this form, the center is given by $(-u, -v, -w)$ and the radius is calculated as $r = \sqrt{u^2 + v^2 + w^2 - d}$.

Step 1: Normalize the equation.
The given equation has a coefficient of 3 for the squared terms. Divide the entire equation by 3 to reach the standard form: \[ x^{2} + y^{2} + z^{2} + 4x + 6y + 8z + 10 = 0 \]

Step 2: Identify the center.
Compare the normalized equation with $x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d = 0$: - $2u = 4 \Rightarrow u = 2$ - $2v = 6 \Rightarrow v = 3$ - $2w = 8 \Rightarrow w = 4$ - $d = 10$ The center $(-u, -v, -w)$ is therefore $(-2, -3, -4)$.

Step 3: Calculate the radius.
Using the radius formula: \[ r = \sqrt{(2)^2 + (3)^2 + (4)^2 - 10} \] \[ r = \sqrt{4 + 9 + 16 - 10} = \sqrt{19} \] Thus, the center is $(-2, -3, -4)$ and the radius is $\sqrt{19}$.