If the equation of a circle is $a x^{2} + (2 a - 3) y^{2} - 4 x - 1 = 0$ then its centre is

(2, 0)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(\frac{2}{3}, 0)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

(- \frac{2}{3}, 0)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

none of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D $(\frac{2}{3}, 0)$
Given equation of circle is
$a x^{2} + (2 a - 3) y^{2} - 4 x - 1 = 0$
Since, it represents a circle ,
Coefficient of $x^{2}$ =Coefficient of $y^{2}$
$a = 2 a - 3$
$\Rightarrow a = 3$
Hence, the equation of circle is
$3 x^{2} + 3 y^{2} - 4 x - 1 = 0$
$\Rightarrow x^{2} + y^{2} - \frac{4}{3} x - \frac{1}{3} = 0$
So, center is at $(\frac{2}{3}, 0)$

Suggest Corrections

Similar questions

a + b + c + d + e

a x^{2} + b y^{2} + c x + d y + e = 0

(\frac{4}{5}, 0)

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

3 \sqrt{2}

2 \sqrt{3}

2 \sqrt{2}

Join BYJU'S Learning Program