Question:
If (a, b) is the common point of the circles \(x^2 + y^2 - 4x + 4y - 1 = 0\) and \(x^2 + y^2 + 2x - 4y + 1 = 0\), then \(a^2 + b^2 =\)
If (a, b) is the common point of the circles \(x^2 + y^2 - 4x + 4y - 1 = 0\) and \(x^2 + y^2 + 2x - 4y + 1 = 0\), then \(a^2 + b^2 =\)
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Use elimination of like terms between circles to find the point of intersection.
Updated On: Jun 4, 2025
- \(\dfrac{1}{5}\)
- \(5\)
- \(25\)
- \(\dfrac{1}{25}\)
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The Correct Option is A
Solution and Explanation
Subtract equations to eliminate \(x^2 + y^2\):
\[ (-4x + 4y -1) - (2x - 4y + 1) = -6x + 8y -2 = 0 \Rightarrow 3x - 4y = 1 \Rightarrow x = \frac{4y + 1}{3} \] Substitute into any original circle, solve for y, then get x, compute \(a^2 + b^2\)
\[ (-4x + 4y -1) - (2x - 4y + 1) = -6x + 8y -2 = 0 \Rightarrow 3x - 4y = 1 \Rightarrow x = \frac{4y + 1}{3} \] Substitute into any original circle, solve for y, then get x, compute \(a^2 + b^2\)
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