Question

If the equation of a circle is λx² + (2λ − 3) y² − 4x + 6y − 1 = 0, then the coordinates of centre are
(a) (4/3, −1)
(b) (2/3, −1)
(c) (−2/3, 1)
(d) (2/3, 1)

Answer 1

(b) $\left(\frac{2}{3},-1\right)$

To find the centre:
Coefficient of x² = Coefficient of y²
$\therefore \lambda =2\lambda -3\Rightarrow \lambda =3$

Therefore, the given equation can be rewritten as $3x^2+3y^2-4x+6y-1=0$.
$\therefore x^2+y^2-\frac{4}{3}x+2y-\frac{1}{3}=0$

Thus, the coordinates of the centre is $\left(\frac{2}{3},-1\right)$.