Home > Coordinate Geometry > jee_main_2026_april_4_shift_1 > 19f9f5cb-d444-4f69-b5e7-90ff7ef73e27

Not the exact question you were looking for?

Paste your question to our Mathbee AI Mentor below to get an instant step-by-step solution.

Coordinate Geometry

Circles, Center and Radius

jee_main_2026_april_4_shift_1

Grade 12

Question:

The center of the circle x^2 + y^2 - 4x + 6y - 12 = 0 is:

A. (2, -3)

B. (-2, 3)

C. (2, 3)

D. (-2, -3)

Step-by-Step Solution

Key Concept: General equation: x^2 + y^2 + 2gx + 2fy + c = 0, center = (-g, -f).

Step 1: Center = (2, -3).

Correct Answer: A

Mathbee AI Mentor (Free Demo)

Confused by the solution? Ask the AI to explain a specific step, tell you where you went wrong, or break down the key trap in this question.