The circle concentric with $x^{2} + y^{2} + 4 x + 6 y + 3 = 0$ and radius $2$ is

x^{2} + y^{2} + 4 x + 6 y + 9 = 0

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

x^{2} + y^{2} + 4 x + 6 y + 9 =

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

x^{2} + y^{2} - 4 x - 6 y + 9 = 0

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

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

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

Open in App

Solution

The correct option is B $x^{2} + y^{2} + 4 x + 6 y + 9 =$
Equation of the circle is given as $x^{2} + y^{2} + 4 x + 6 y + 3 = 0$
$x^{2} + 4 x + 4 + y^{2} + 6 y + 9 = - 3 + 9 + 4$
$(x^{2} + 2)^{2} (y + 3)^{2} = 10$
$∴$ Centre of the circle is $(- 2, - 3)$
Circle with centre $(- 2, 3)$ and radius $2$
$(x + 2)^{2} + (y + 3)^{2} = (2)^{2}$
$x^{2} + 4 x + 4 + y^{2} + 6 x + 9 = 4$
$x^{2} + y^{2} + 4 x + 6 x + 9 = 0$ .

Suggest Corrections

Similar questions

P

x^{2} + y^{2} + 4 x - 6 y + 9 {sin}^{2} α + 13 {cos}^{2} α = 0

2 α .

P

x^{2} + y^{2} - 6 x - 4 y = 0

2^{nd}

4^{th}

x^{2} + y^{2} + 8 x + 10 y - 7 = 0

x^{2} + y^{2} - 4 x - 6 y = 0

x^{2} + y^{2} - 4 x + 6 y + 8 = 0

x^{2} + y^{2} - 10 x - 6 y + 14 = 0

(x - 2)^{2} + (y + 3)^{2} = 16

Join BYJU'S Learning Program