Question

If $2x^2+\lambda xy+2y^2+(\lambda -4)x+6y-5=0$
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

Answer: (D)

(None of these)

The given equation is $2x^2+\lambda xy+2y^2+(\lambda -4)x+6y-5=0$,
which can be rewritten as

$x^2+\frac{\lambda xy}{2}+y^2+\frac{(\lambda -4)}{2}x+3y-\frac{5}{2}=0$......(i)

Comparing the given equation with $x^2+y^2+2gx+2fy+c=0$....(ii)
we get: $\lambda =0$ as equation (ii) contains no $xy$ term

Substitute $\lambda =0$ in equation (i), we get

$\therefore x^2+y^2-2x+3y-\frac{5}{2}=0$ .....(iii)

Compare equation(iii) with equation(ii), we get $g=-1,f=\frac{3}{2}$ and $c=-\frac{5}{2}$

$\therefore$ Radius$=\sqrt{g^2+f^2-c}$
$=\sqrt{(-1{)}^2+{\left(\frac{3}{2}\right)}^2+\frac{5}{2}}$

$=\sqrt{1+\frac{9}{4}+\frac{5}{2}}$
$=\sqrt{\frac{4+9+10}{4}}$
$=\sqrt{\frac{23}{4}}$
$=\frac{\sqrt{23}}{2}$