lf 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 $\frac{1}{c} =$

a^{2} + b^{2}

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\frac{1}{a} + \frac{1}{b}

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

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a + b

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Solution

The correct option is D $\frac{1}{a^{2}} + \frac{1}{b^{2}}$
$S_{1} : x^{2} + y^{2} + 2 a x + c = 0 \Rightarrow C_{1} \equiv (- a, 0)$
$S_{2} : x^{2} + y^{2} + 2 b y + c = 0 \Rightarrow C_{2} \equiv (0, - b)$
$∴$ Equation of common tangent is $S_{2} - S_{1} = 0$
$2 b y - 2 a x = 0$
$b y = a x$ _(1)
The equation of line passing through centres of circles is given by
$y - 0 = \frac{(- b)}{a} (x + a)$
$a y + b x = - a b$ __(2)
From (1) & (2)
$x = - \frac{a b^{2}}{a^{2} + b^{2}}$ & $y = - \frac{a^{2} b}{a^{2} + b^{2}}$
$x^{2} + y^{2} + 2 a x + c = 0$
$\Rightarrow \frac{a^{2} b^{2}}{(a^{2} + b^{2})^{2}} (a^{2} + b^{2}) - \frac{2 a^{2} b^{2}}{a^{2} + b^{2}} + c = 0$
$\Rightarrow c = \frac{a^{2} b^{2}}{a^{2} + b^{2}}$
$\Rightarrow \frac{1}{c} = \frac{1}{a^{2}} + \frac{1}{b^{2}}$
Hence, option 'C' is correct.

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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^{2} = 0

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

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

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 + 2 b y + 1 = 0

x^{2} + y^{2} + 2 b x + 2 a y + 1 = 0

{(a + b)}^{2}

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