The circles whose equations are $x^{2} + y^{2} + c^{2} = 2 a x$ and $x^{2} + y^{2} + c^{2} - 2 b y = 0$ will touch one another externally, if

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$\frac{1}{b^{2}} + \frac{1}{c^{2}} = \frac{1}{a^{2}}$

none of these

$\frac{1}{c^{2}} + \frac{1}{a^{2}} = \frac{1}{b^{2}}$

$\frac{1}{a^{2}} + \frac{1}{b^{2}} = \frac{1}{c^{2}}$

answer is C.

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Detailed Solution

The two circles are

$x^{2} + y^{2} - 2 a x + c^{2} = 0$ and $x^{2} + y^{2} - 2 b y + c^{2} = 0$

Centres and radii of these two circles are

Centres $\begin{array}{cc} C_{1} (a, 0) & C_{2} (0, b) \\ r_{1} = \sqrt{a^{2} - c^{2}} & r_{2} = \sqrt{b^{2} - c^{2}} \end{array}$

Radii:

Given that the two circles touch each other externally.

$\begin{array}{l} ∴ C_{1} C_{2} = r_{1} + r_{2} \\ \Rightarrow \sqrt{a^{2} + b^{2}} = \sqrt{a^{2} - c^{2}} + \sqrt{b^{2} - c^{2}} \\ \Rightarrow a^{2} + b^{2} = a^{2} - c^{2} + b^{2} - c^{2} + 2 \sqrt{a^{2} - c^{2}} \sqrt{b^{2} - c^{2}} \\ \Rightarrow c^{4} = a^{2} b^{2} - c^{2} (a^{2} + b^{2}) + c^{4} \\ \Rightarrow a^{2} b^{2} = c^{2} (a^{2} + b^{2}) \Rightarrow \frac{1}{a^{2}} + \frac{1}{b^{2}} = \frac{1}{c^{2}} \end{array}$