Question

The equation of the circle passing through the points (1, 1), (2, 2), and (3, 1) is

• A. \( x^2 + y^2 - 2x - 2y + 1 = 0 \)
• B. \( x^2 + y^2 - 4x - 2y + 5 = 0 \)
• C. \( x^2 + y^2 - 4x - 4y + 7 = 0 \)
• D. \( x^2 + y^2 - 2x - 4y + 3 = 0

Hint:

\textbf{Key Fact:} Use the general circle equation and solve the system of equations formed by substituting the given points.

Answer 1

The Correct Option is D

• General Equation of Circle: The equation of a circle is \( x^2 + y^2 + 2gx + 2fy + c = 0 \). We need to find \( g \), \( f \), and \( c \) using the given points.
• Substitute Points:
  • (1, 1): \( 1^2 + 1^2 + 2g(1) + 2f(1) + c = 0 \Rightarrow 1 + 1 + 2g + 2f + c = 0 \Rightarrow 2g + 2f + c = -2 \).
  • (2, 2): \( 2^2 + 2^2 + 2g(2) + 2f(2) + c = 0 \Rightarrow 4 + 4 + 4g + 4f + c = 0 \Rightarrow 4g + 4f + c = -8 \).
  • (3, 1): \( 3^2 + 1^2 + 2g(3) + 2f(1) + c = 0 \Rightarrow 9 + 1 + 6g + 2f + c = 0 \Rightarrow 6g + 2f + c = -10 \).
• Solve Equations:
  • Subtract equation (1) from (2): \[ (4g + 4f + c) - (2g + 2f + c) = -8 - (-2) \Rightarrow 2g + 2f = -6 \Rightarrow g + f = -3. \]
  • Subtract equation (1) from (3): \[ (6g + 2f + c) - (2g + 2f + c) = -10 - (-2) \Rightarrow 4g = -8 \Rightarrow g = -2. \]
  • Substitute \( g = -2 \) into \( g + f = -3 \): \[ -2 + f = -3 \Rightarrow f = -1. \]
  • Substitute \( g = -2 \), \( f = -1 \) into equation (1): \[ 2(-2) + 2(-1) + c = -2 \Rightarrow -4 - 2 + c = -2 \Rightarrow c = 4. \]
• Circle Equation: Substitute values back: \[ x^2 + y^2 + 2(-2)x + 2(-1)y + 4 = 0 \Rightarrow x^2 + y^2 - 4x - 2y + 4 = 0. \]
• Conclusion: The correct equation is \( \boxed{x^2 + y^2 - 4x - 2y + 4 = 0} \).