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For the equation 2x² + (lambda)xy + 2y² + (lambda - 4)x + 6y - 5 = 0 to represent a circle (for a suitable lambda), find its centre and radius.
- Centre (1, -3/2), radius sqrt(53)/2
- Centre (-1, 3/2), radius sqrt(53)/2
- Centre (1, -3/2), radius sqrt(53)/4
- Centre (-1, -3/2), radius 7/2
Correct answer: Centre (1, -3/2), radius sqrt(53)/2
Solution
Requiring no xy term gives lambda = 0; the equation becomes 2x² + 2y² - 4x + 6y - 5 = 0, i.e. x² + y² - 2x + 3y - 5/2 = 0, giving centre (1, -3/2) and radius sqrt(1 + 9/4 + 5/2) = sqrt(53)/2.
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