Circles — Easy JEE math MCQ
The center of the circle $x^2 + y^2 - 4x + 6y - 12 = 0$ is:
- A. $(2, -3)$
- B. $(-2, 3)$
- C. $(4, -6)$
- D. $(-4, 6)$
Solution
Compare with $x^2 + y^2 + 2gx + 2fy + c = 0$
$2g = -4 \Rightarrow g = -2$
$2f = 6 \Rightarrow f = 3$
Center $= (-g, -f) = (2, -3)$