Question

For any real number \(\lambda \neq 1\), the centre of the circle that passes through \(A(1, \lambda)\), \(B(\lambda, 1)\), and \(C(\lambda, \lambda)\) is

• A. \(\left(\frac{1+\lambda}{2}, \frac{1+\lambda}{2}\right)\)
• B. \(\left(\frac{1+2\lambda}{3}, \frac{1+2\lambda}{3}\right)\)
• C. \((1+2\lambda, 1+2\lambda)\)
• D. \(\left(\frac{\lambda}{2}, \frac{\lambda}{2}\right)\)

Hint:

For a circle through three points, substitute points into general circle equation and solve for the center by eliminating the radius.

Answer 1

The Correct Option is A

Step 1: General circle equation.
Let the circle be \((x - h)^2 + (y - k)^2 = r^2\) with center \((h,k)\).

Step 2: Substitute points.
Substitute \(A(1,\lambda)\), \(B(\lambda,1)\), \(C(\lambda,\lambda)\) to get three equations:
\((1-h)^2 + (\lambda - k)^2 = r^2\), \((\lambda - h)^2 + (1 - k)^2 = r^2\), \((\lambda - h)^2 + (\lambda - k)^2 = r^2\)

Step 3: Solve for h and k.
Subtract third equation from first and second:
\((1-h)^2 + (\lambda - k)^2 - [(\lambda - h)^2 + (\lambda - k)^2] = 0 \implies h = \frac{1+\lambda}{2}\)
\((\lambda - h)^2 + (1-k)^2 - [(\lambda - h)^2 + (\lambda - k)^2] = 0 \implies k = \frac{1+\lambda}{2}\)

Step 4: Final conclusion.
Hence, the center of the circle is \[ \boxed{\left(\frac{1+\lambda}{2}, \frac{1+\lambda}{2}\right)} \]