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The condition that the circles which passes through the points (0, a), (0, –a) and touch the line y = mx + c will cut orthogonally is

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$c^{2} = a^{2} (1 + m^{2})$

$c^{2} = a^{2} (2 + m^{2})$

$c^{2} = a^{2} (3 + m^{2})$

$c^{2} = a^{2} (4 + m^{2})$

answer is B.

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Detailed Solution

$\begin{array}{rcl} c & = & (0, 0) r = a \\ x^{2} + y^{2} & = & a^{2} touches the \\ y & = & mx + c \\ ∴ r & = & d \\ a & = & \frac{|c|}{\sqrt{m^{2} + 1}} \\ a^{2} (1 + m^{2}) & = & c^{2} \end{array}$

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