Question

If two circles which pass through the points \( (0, a) \) and \( (0, -a) \) and touch the line \( y = mx + c \) cut orthogonally, then:

• A. \( c^2 = a^2(1 + m^2) \)
• B. \( c^2 = a^2(2 + m^2) \)
• C. \( c^2 = 2a^2(1 + 2m^2) \)
• D. \( 2c^2 = a^2(1 + m^2) \)

Hint:

For problems involving circles, use the standard circle equation and apply given conditions step by step.

Answer 1

The Correct Option is B

1. The equation of a circle passing through \((0, a)\) and \((0, -a)\) is:

\[ x^2 + y^2 + 2gx + 2fy + c = 0. \]

2. Since the circles pass through \((0, a)\) and \((0, -a)\):

\[ f = 0, \quad c = -a^2. \]

3. The equation simplifies to:

\[ x^2 + y^2 + 2gx + c = 0. \]

4. If the circles touch the line \(y = mx + c\) orthogonally, the condition for orthogonality is:

\[ c^2 = a^2(2 + m^2). \]

Thus, the correct answer is \(c^2 = a^2(2 + m^2)\).