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If 2x2 + λXy + 2y2 + (λ − 4) X + 6y − 5 = 0 is the Equation of a Circle, Then Its Radius is - Mathematics

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प्रश्न

If 2x2 + λxy + 2y2 + (λ − 4) x + 6y − 5 = 0 is the equation of a circle, then its radius is

विकल्प

  • \[3\sqrt{2}\]

  • \[2\sqrt{3}\]

  • \[2\sqrt{2}\]

  • none of these

MCQ
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उत्तर

none of these
The given equation is 2x2 + λxy + 2y2 + (λ − 4) x + 6y − 5 = 0, which can be rewritten as \[x^2 + \frac{\lambda xy}{2} + y^2 + \frac{\left( \lambda - 4 \right)}{2}x + 3y - \frac{5}{2} = 0\].

Comparing the given equation with \[x^2 + y^2 + 2gx + 2fy + c = 0\] we get: \[\lambda = 0\]

∴ \[x^2 + y^2 - 2x + 3y - \frac{5}{2} = 0\]

∴ Radius = \[\sqrt{\left( - 1 \right)^2 + \left( \frac{3}{2} \right)^2 + \frac{5}{2}} = \sqrt{1 + \frac{9}{4} + \frac{5}{2}} = \sqrt{\frac{23}{4}} = \frac{\sqrt{23}}{2}\]

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Advanced Concept of Circle - Standard Equation of a Circle
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 24: The circle - Exercise 24.6 [पृष्ठ ३९]

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आरडी शर्मा Mathematics [English] Class 11
अध्याय 24 The circle
Exercise 24.6 | Q 2 | पृष्ठ ३९

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