Question

If 2x² + λxy + 2y² + (λ − 4) x + 6y − 5 = 0 is the equation of a circle, then its radius is
(a) $3\sqrt{2}$
(b) $2\sqrt{3}$
(c) $2\sqrt{2}$
(d) none of these

Answer 1

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