The equation of the sphere through the circle `x^(2)+y^(2)+z^(2)=9,2x+3y+4z=5` and the point `(1,2,3)` is …....

Answer

To find the equation of the sphere that passes through the circle defined by the equations \(x^2 + y^2 + z^2 = 9\) and \(2x + 3y + 4z = 5\), and also passes through the point \((1, 2, 3)\), we can follow these steps:

### Step 1: Write the general equation of the sphere
The general equation of a sphere can be expressed as:
\[ x^2 + y^2 + z^2 + ax + by + cz + d = 0 \]
Since the sphere passes through the circle defined by \(x^2 + y^2 + z^2 = 9\) and \(2x + 3y + 4z = 5\), we can rewrite the equation as:
\[ x^2 + y^2 + z^2 - 9 + \lambda(2x + 3y + 4z - 5) = 0 \]
where \(\lambda\) is a parameter.

### Step 2: Substitute the point \((1, 2, 3)\) into the equation
Now, we need to substitute the coordinates of the point \((1, 2, 3)\) into the equation to find \(\lambda\):
\[ 1^2 + 2^2 + 3^2 - 9 + \lambda(2(1) + 3(2) + 4(3) - 5) = 0 \]
Calculating each term:
\[ 1 + 4 + 9 - 9 + \lambda(2 + 6 + 12 - 5) = 0 \]
This simplifies to:
\[ 5 + \lambda(15) = 0 \]
Thus:
\[ \lambda = -\frac{5}{15} = -\frac{1}{3} \]

### Step 3: Substitute \(\lambda\) back into the sphere equation
Now, we substitute \(\lambda = -\frac{1}{3}\) back into the equation of the sphere:
\[ x^2 + y^2 + z^2 - 9 - \frac{1}{3}(2x + 3y + 4z - 5) = 0 \]
Multiplying through by 3 to eliminate the fraction:
\[ 3(x^2 + y^2 + z^2 - 9) - (2x + 3y + 4z - 5) = 0 \]
Expanding this gives:
\[ 3x^2 + 3y^2 + 3z^2 - 27 - 2x - 3y - 4z + 5 = 0 \]
Combining like terms:
\[ 3x^2 + 3y^2 + 3z^2 - 2x - 3y - 4z - 22 = 0 \]

### Final Equation of the Sphere
Thus, the equation of the sphere is:
\[ 3x^2 + 3y^2 + 3z^2 - 2x - 3y - 4z - 22 = 0 \]

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