The equation of a diameter of circle $x^{2} + y^{2} - 6 x + 2 y = 0$ passing through origin is

x + 3 y = 0

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

x - 3 y = 0

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

3 x + y = 0

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

3 x - y = 0

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

Open in App

Solution

The correct option is D $3 x + y = 0$
Comparing with standard eqn of circle $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ ,
$g = - 3$
$f = 1$
So, the centre $M$ of the circle is at $(- g, - f) i . e . (3, - 1)$
Also it has been stated that the circle passes through the origin $O (0, 0)$
So the diameter is $O M$ straight line.
The equation of diameter is $\frac{(y - 0)}{(x - 0)} = \frac{(3 - 0)}{(- 1 - 0)}$
$y + 3 x = 0$

Suggest Corrections

Similar questions

x^{2} + y^{2} - 6 x + 2 y - 8 = 0

x^{2} + y^{2} - 2 x + 4 y = 0

3 x + y + 5 = 0

x^{2} + y^{2} = 16

3 x^{2} + 3 y^{2} - 2 x + 12 y - 9 = 0

x^{2} + y^{2} + 6 x + 2 y - 15 = 0

x^{2} + y^{2} + 2 x + 3 y + 1 = 0, x^{2} + y^{2} + x + y + 2 = 0, x^{2} + y^{2} + 3 x + 2 y + 1 = 0

Join BYJU'S Learning Program