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. (1,2)

Answer 1

The correct option is C

(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).