Question

If the equation of a circle is $\lambda x^2+(2\lambda -3)y^2-4x+6y-1=0$
then the coordinates of centre are

• A. $(\frac{4}{3},-1)$
• B. $(\frac{2}{3},-1)$
• C. $(-\frac{2}{3},1)$
• D. $(\frac{2}{3},1)$

Answer 1

The correct option is B

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

To find the centre:

coefficient of $x^2$.= coefficient of $y^2$

$\therefore \lambda =2\lambda -3\Rightarrow \lambda =3$

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