Consider the circles $x^{2} + y^{2} + 2 a x + c = 0$ and $x^{2} + y^{2} + 2 b y + c = 0$ . The two circles touch each other if.

c = \sqrt{a^{2} + b^{2}}

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

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

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

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Solution

The correct option is B $\frac{1}{c} = \frac{1}{a^{2}} + \frac{1}{b^{2}}$
let , $S_{1} = x^{2} + y^{2} + 2 a x + c$ , center = $(a, 0)$
$S_{2} = x^{2} + y^{2} + 2 b y + c$ , center = $(0, b)$

we know that circles touches each other,

sun of radii is equal to distance between their centers

$r_{1} + r_{2} = \sqrt{a^{2} + b^{2}}$

$\Rightarrow \sqrt{a^{2} - c} + \sqrt{b^{2} - c} = \sqrt{a^{2} + b^{2}}$

we can have ,

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

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The two circles $x^{2} + y^{2} + 2 a x + c = 0 a n d x^{2} + y^{2} + 2 b y + c = 0$ touch if $\frac{1}{a^{2}} + \frac{1}{b^{2}} =$

If the circles $x^{2} + y^{2} + 2 a x + c = 0$ and $x^{2} + y^{2} + 2 b y + c = 0$ touch each other, then

x^{2} + y^{2} + 2 a x + c = 0

x^{2} + y^{2} + 2 b y + c = 0

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

c_{1}, c_{2}

r_{1}, r_{2}

c_{1} + c_{2} = r_{1} \pm r_{2}

x^{2} + y^{2} + 2 a x + c^{2} = 0

x^{2} + y^{2} + 2 b y + c^{2} = 0

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

\in R^{+}

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} - \frac{z^{2}}{c^{2}} = 1

\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1

- \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1

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