Question

The centre of the circle (x)${}^2$+(y)${}^2$ – 4x – 6y – 12 = 0 is

• A. (2, -3)
• B. (-2 , 3)
• C. (-2, -3)
• D. (2,3)
• E. (1,2)

Answer 1

The correct option is D

(2,3)

The equation becomes -- (x-2)${}^2$+(y-3)${}^2$ = 5${}^2$

$\therefore$The centre of the circle x${}^2$+y${}^2$ - 4x - 6y - 12 = 0 is (2, 3).