Question

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

Answer 1

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