The limiting points of the system of coaxial circles $x^{2} + y^{2} + 2 f y + 25 = 0$ are

(0, \pm 25)

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(0, \pm 5)

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(\pm 5, 0)

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Not existing

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Solution

The correct option is A $(0, \pm 5)$
Centre of the given coaxial system of circle is, $(0, f)$
and radius is $= \sqrt{f^{2} - 25} = 0$ for limiting points
$\Rightarrow f = \pm 5$ . Hence limiting points are $(0, \pm 5)$

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