Find the center of circle $x^{2} + y^{2} - 6 x - 8 y - 24 = 0$

Open in App

Solution

$x^{2} + y^{2} - 6 x - 8 y - 24 = 0 x^{2} - 6 x + 9 + y^{2} - 8 y + 16 = 49 (x - 3)^{2} + (y - 4)^{2} = 49$
So the center is $(3, 4)$

Suggest Corrections

Similar questions

x^{2} + y^{2} - 6 x - 8 y + 24 = 0

The point P(9, 12) lies inside the circle $x^{2} + y^{2} - 6 x - 8 y - 24 = 0$

x^{2} + y^{2} - 6 x + 4 y - 23 = 0

x^{2} + y^{2} + 6 x - 8 y - 24 = 0

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

x^{2} + y^{2} - 6 x - 8 y - 24 = 0

Join BYJU'S Learning Program