The plane `2x-2y+z+12=0` touches the sphere `x^(2)+y^(2)+z^(2)-2x-4y+2z-3=0` at the point

Options

A. `(1,-4,-2)`
B. `(-1,4,-2)`
C. `(-1,-4,2)`
D. `(1,4,-2)`

Answer

To find the point where the plane \(2x - 2y + z + 12 = 0\) touches the sphere \(x^2 + y^2 + z^2 - 2x - 4y + 2z - 3 = 0\), we can follow these steps:

### Step 1: Rewrite the equation of the sphere
We start by rewriting the equation of the sphere in standard form. The given equation is:
\[ x^2 + y^2 + z^2 - 2x - 4y + 2z - 3 = 0 \]
We can complete the square for each variable:
- For \(x\): \(x^2 - 2x = (x - 1)^2 - 1\)
- For \(y\): \(y^2 - 4y = (y - 2)^2 - 4\)
- For \(z\): \(z^2 + 2z = (z + 1)^2 - 1\)

Substituting these into the equation gives:
\[ (x - 1)^2 - 1 + (y - 2)^2 - 4 + (z + 1)^2 - 1 - 3 = 0 \]
Simplifying this, we have:
\[ (x - 1)^2 + (y - 2)^2 + (z + 1)^2 - 9 = 0 \]
Thus, the standard form of the sphere is:
\[ (x - 1)^2 + (y - 2)^2 + (z + 1)^2 = 9 \]
This indicates that the center of the sphere is \((1, 2, -1)\) and the radius is \(3\).

### Step 2: Find the normal vector of the plane
The equation of the plane is given as:
\[ 2x - 2y + z + 12 = 0 \]
The normal vector of the plane can be extracted from the coefficients of \(x\), \(y\), and \(z\):
\[ \mathbf{n} = (2, -2, 1) \]

### Step 3: Find the distance from the center of the sphere to the plane
To find the distance \(d\) from the center of the sphere \((1, 2, -1)\) to the plane, we use the formula:
\[ d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \]
where \(A = 2\), \(B = -2\), \(C = 1\), \(D = 12\), and \((x_0, y_0, z_0) = (1, 2, -1)\).

Calculating the numerator:
\[ |2(1) - 2(2) + 1(-1) + 12| = |2 - 4 - 1 + 12| = |9| = 9 \]

Calculating the denominator:
\[ \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 \]

Thus, the distance \(d\) is:
\[ d = \frac{9}{3} = 3 \]

### Step 4: Verify that the distance equals the radius
Since the radius of the sphere is also \(3\), and the distance from the center of the sphere to the plane is \(3\), this confirms that the plane touches the sphere.

### Step 5: Find the point of tangency
The point of tangency lies along the line from the center of the sphere to the plane. We can find the point of tangency by moving from the center of the sphere \((1, 2, -1)\) in the direction of the normal vector \((2, -2, 1)\).

To find the coordinates of the point of tangency, we can use the parametric equation:
\[ P = (1, 2, -1) + t(2, -2, 1) \]
We need to find \(t\) such that the distance moved equals the radius, which is \(3\):
\[ \sqrt{(2t)^2 + (-2t)^2 + (t)^2} = 3 \]
Simplifying:
\[ \sqrt{4t^2 + 4t^2 + t^2} = 3 \implies \sqrt{9t^2} = 3 \implies 3|t| = 3 \implies |t| = 1 \]
Thus, \(t = 1\) or \(t = -1\). We take \(t = 1\) for the point of tangency.

Substituting \(t = 1\):
\[ P = (1 + 2, 2 - 2, -1 + 1) = (3, 0, 0) \]

### Final Answer
The point at which the plane touches the sphere is \((3, 0, 0)\).

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