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  • The circle whose equations are <img alt="\mathrm{x} ^2+\mathrm{y}^ 2+\mathrm{c}^ 2=2 \mathrm{ax}$ and $\mathrm{x} ^2+\mathrm{y}^ 2+\mathrm{c} ^2-2 \mathrm{by}=0" src="/latex-image/?%5Cmathrm%7Bx%7D

The circle whose equations are \mathrm{x} ^2+\mathrm{y}^ 2+\mathrm{c}^ 2=2 \mathrm{ax}$ and $\mathrm{x} ^2+\mathrm{y}^ 2+\mathrm{c} ^2-2 \mathrm{by}=0 will touch one another externally if

Option: 1

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


Option: 2

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


Option: 3

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


Option: 4

None of these


Answers (1)

best_answer

The two circles are
\mathrm{x}^ 2+\mathrm{y} ^2-2 \mathrm{ax}+\mathrm{c}^ 2=0$ and $\mathrm{x}^ 2+\mathrm{y}^ 2-2 \mathrm{by}+\mathrm{c}^ 2=0

\mathrm{Centres} : C1 (\mathrm{a}, 0), \mathrm{C} 2(0, \mathrm{~b})
\mathrm{radii : r_{1}=\sqrt{a^{2}-c^{2}}, r_{2}=\sqrt{b^{2}-c^{2}})}

Since, the two circles touch each other externally, therefore, \mathrm{C1C2 =\mathrm{r} 1+\mathrm{r} 2}

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

\mathrm{\mathrm{c} 4=\mathrm{a} 2 \mathrm{~b} 2-\mathrm{c} 2(\mathrm{a} 2+\mathrm{b} 2)+\mathrm{c} 4}
\mathrm{ \mathrm{a} 2 \mathrm{~b} 2=\mathrm{c} 2(\mathrm{a} 2+\mathrm{b} 2) \Rightarrow \frac{1}{\mathrm{a}^{2}}+\frac{1}{b^{2}}=\frac{1}{c^{2}}}

Posted by

Ajit Kumar Dubey

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