The equation of a circle which passes through the points of intersection of the circles \[ 2x^2 + 2y^2 - 2x + 6y - 3 = 0, \quad x^2 + y^2 + 4x + 2y + 1 = 0 \] and whose centre lies on the common chord of these circles is:
The equation of a circle which passes through the points of intersection of the circles \[ 2x^2 + 2y^2 - 2x + 6y - 3 = 0, \quad x^2 + y^2 + 4x + 2y + 1 = 0 \] and whose centre lies on the common chord of these circles is:
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- \( 2x^2 + 2y^2 - 3x + 4y - 2 = 0 \)
- \( x^2 + y^2 + 2x + 5y - 2 = 0 \)
- \( 3x^2 + 3y^2 - 2x + 4y - 3 = 0 \)
\( 4x^2 + 4y^2 + 6x + 10y - 1 = 0 \)
The Correct Option is D
Solution and Explanation
Step 1: Finding the Equation of the Common Chord
The given circles are: \[ 2x^2 + 2y^2 - 2x + 6y - 3 = 0, \] \[ x^2 + y^2 + 4x + 2y + 1 = 0. \] Subtracting the second equation from the first: \[ (2x^2 + 2y^2 - 2x + 6y - 3) - (x^2 + y^2 + 4x + 2y + 1) = 0. \] \[ x^2 + y^2 - 6x + 4y - 4 = 0. \]
Step 2: Finding the Required Circle
A circle passing through the intersection of these two given circles is of the form: \[ S_1 + \lambda S_2 = 0. \] Substituting and simplifying using the condition that the center lies on the common chord, we derive: \[ 4x^2 + 4y^2 + 6x + 10y - 1 = 0. \]
Step 3: Conclusion
Thus, the final answer is: \[ \boxed{4x^2 + 4y^2 + 6x + 10y - 1 = 0}. \]
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Given the vectors:
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\[ \mathbf{a} \cdot \mathbf{x} = 3 \]
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