Question:medium

The radical plane of the spheres \(x^2+y^2+z^2+4x-2y+2z+6=0\) and \(x^2+y^2+z^2+2x-4y-2z+6=0\) is

Show Hint

To find the radical plane of two spheres, subtract their equations and simplify.
  • \(x-y+2z=0\)
  • \(x+y+2z=0\)
  • \(x+y-2z=0\)
  • \(x-y-2z=0\)
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
The radical plane of two spheres is the locus of points from which the lengths of the tangents to the two spheres are equal. Its equation is found by simply subtracting the equations of the two spheres.

Step 2: Key Formula or Approach:

Let the equations of the two spheres be $S_1 = 0$ and $S_2 = 0$, where the coefficients of $x^2, y^2, z^2$ are unity.
$S_1 : x^2+y^2+z^2+2u_1x+2v_1y+2w_1z+d_1=0$
$S_2 : x^2+y^2+z^2+2u_2x+2v_2y+2w_2z+d_2=0$
The equation of the radical plane is given by $S_1 - S_2 = 0$.

Step 3: Detailed Explanation:

We are given the two sphere equations:
$S_1 : x^2+y^2+z^2+4x-2y+2z+6=0$
$S_2 : x^2+y^2+z^2+2x-4y-2z+6=0$
To find the radical plane, we compute $S_1 - S_2 = 0$:
$(x^2+y^2+z^2+4x-2y+2z+6) - (x^2+y^2+z^2+2x-4y-2z+6) = 0$
The quadratic terms ($x^2, y^2, z^2$) cancel out:
$(4x-2x) + (-2y - (-4y)) + (2z - (-2z)) + (6-6) = 0$
$2x + (-2y+4y) + (2z+2z) + 0 = 0$
$2x + 2y + 4z = 0$
This is the equation of the radical plane. We can simplify it by dividing the entire equation by 2:
$x + y + 2z = 0$
This matches option (B).

Step 4: Final Answer:

The equation of the radical plane is $x+y+2z=0$.
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