Calculate the center of a circle from the equation: $x^{2} + y^{2} - 4 x + 6 y + 24 = 0$

(2, 3)

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

(- 2, - 3)

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

(2, - 3)

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

(- 2, 3)

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

Open in App

Solution

The correct option is C $(2, - 3)$
The general form of a circle is $(x - h)^{2} + (y - k)^{2} - r^{2} = 0$
Compare the general form with the given equation $x^{2} + y^{2} - 4 x + 6 y + 24 = 0$
So, $- 4 = - 2 h; 6 = - 2 k$
$h = 2, k = - 3$
So, the center of the circle is $(2, - 3)$ .

Suggest Corrections

Similar questions

(1, 1)

x^{2} + y^{2} + 4 x - 6 y - 3 = 0

(2, 3)

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

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

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

(- 2, 3)

x^{2} + y^{2} - 4 x - 6 y + 5 = 0

(- 3, - 1), (- 2, 3)

(2, 3)

x^{2} + y^{2} + 4 x - 2 y + 3 = 0

x^{2} + y^{2} - 4 x - 6 y + 5 = 0.

Join BYJU'S Learning Program