Question

The equation of a circle that passes through the origin and cut off intercepts $-2$ and $3$ on the X-axis and Y-axis respectively is

• A. $x^2 + y^2 - 2x + 3y = 0$
• B. $x^2 + y^2 + 2x + 3y = 0$
• C. $x^2 + y^2 + 2x - 3y = 0$
• D. $x^2 + y^2 - 2x - 3y = 0$

Hint:

For any circle passing through the origin and cutting intercepts $a$ and $b$ on the axes, its equation can be directly written as $x^2 + y^2 - ax - by = 0$. Here, $a = -2$ and $b = 3$, giving $x^2 + y^2 - (-2)x - (3)y = 0 \Rightarrow x^2 + y^2 + 2x - 3y = 0$ instantly!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
Find the equation of a circle passing through three points: (0,0), (-2,0), and (0,3).

Step 2: Key Formula or Approach:
Use the general circle equation x² + y² + 2gx + 2fy + c = 0. Substitute each coordinate to solve for the constants g, f, and c.

Step 3: Detailed Explanation:
Plugging (0,0): 0 + 0 + 0 + 0 + c = 0 → c = 0. Equation reduces to x² + y² + 2gx + 2fy = 0. Plugging (-2,0): 4 + 0 - 4g + 0 = 0 → g = 1. Plugging (0,3): 0 + 9 + 0 + 6f = 0 → f = -3/2. Substituting back: x² + y² + 2x - 3y = 0.

Step 4: Final Answer:
The circle equation is x² + y² + 2x - 3y = 0, option (C).