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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
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The two circles touch each other if
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Verified Answer
The correct answer is:
$\frac{1}{c}=\frac{1}{a^{2}}+\frac{1}{b^{2}}$
Two circles touch each other, iff distance between two centres $=$ Sum of radius of two circles
$\sqrt{a^{2}+b^{2}}=\sqrt{a^{2}-c}+\sqrt{b^{2}-c}$
On squaring both sides, we get
$a^{2}+b^{2}=a^{2}-c+b^{2}-c+2 \sqrt{\left(a^{2}-c\right)\left(b^{2}-c\right)}$
$\Rightarrow \mathrm{c}=\sqrt{\left(\mathrm{a}^{2}-\mathrm{c}\right)\left(\mathrm{b}^{2}-\mathrm{c}\right)}$
Again, squaring both sides, we get
$$
\mathrm{c}^{2}=\mathrm{a}^{2} \mathrm{~b}^{2}-\mathrm{a}^{2} \mathrm{c}-\mathrm{b}^{2} \mathrm{c}+\mathrm{c}^{2}
$$
$\Rightarrow a^{2} b^{2}=\left(a^{2}+b^{2}\right) c \Rightarrow \frac{1}{c}=\frac{1}{a^{2}}+\frac{1}{b^{2}}$
$\sqrt{a^{2}+b^{2}}=\sqrt{a^{2}-c}+\sqrt{b^{2}-c}$
On squaring both sides, we get
$a^{2}+b^{2}=a^{2}-c+b^{2}-c+2 \sqrt{\left(a^{2}-c\right)\left(b^{2}-c\right)}$
$\Rightarrow \mathrm{c}=\sqrt{\left(\mathrm{a}^{2}-\mathrm{c}\right)\left(\mathrm{b}^{2}-\mathrm{c}\right)}$
Again, squaring both sides, we get
$$
\mathrm{c}^{2}=\mathrm{a}^{2} \mathrm{~b}^{2}-\mathrm{a}^{2} \mathrm{c}-\mathrm{b}^{2} \mathrm{c}+\mathrm{c}^{2}
$$
$\Rightarrow a^{2} b^{2}=\left(a^{2}+b^{2}\right) c \Rightarrow \frac{1}{c}=\frac{1}{a^{2}}+\frac{1}{b^{2}}$
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