Question

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

पर्याय

• (4/3, −1)

• (2/3, −1)

• (−2/3, 1)

• (2/3, 1)

Answer 1

(2/3, −1)

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

\[3 x^2 + 3 y^2 - 4x + 6y - 1 = 0\].

\[\therefore x^2 + y^2 - \frac{4}{3}x + 2y - \frac{1}{3} = 0\]

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