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Number of Common Tangents to Two Circles in Different Conditions

lf the circle...

Question

lf the circles $x^{2} + y^{2} + 2 x + c = 0$ and $x^{2} + y^{2} + 2 y + c = 0$ touch each other then $c =$

\frac{1}{2}

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\frac{1}{4}

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2

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4

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Solution

The correct option is D $\frac{1}{2}$
From point (1),
$S_{1} : x^{2} + y^{2} + 2 x + c = 0 \Rightarrow C_{1} \equiv (- 1, 0)$ & $r_{1} = \sqrt{1 - c}$
$S_{1} : x^{2} + y^{2} + 2 y + c = 0 \Rightarrow C_{2} \equiv (0, - 1)$ & $r_{2} = \sqrt{1 - c}$
case - I
Circles touch each other externally,
$C_{1} C_{2} = r_{1} + r_{2}$
$\Rightarrow \sqrt{1 + 1} = 2 \sqrt{1 - c}$
$2 = 4 - 4 c$
$4 c = 2$
$c = \frac{1}{2}$
case - II
circle touch each other internally.
$C_{1} C_{2} = r_{1} + r_{2}$
$\sqrt{2} = C_{1} C_{2} = 0$
X not possible.

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