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  • If two circles <img alt="\mathrm{x^2+y^2+2 g_1 x+2 f_1 y+2 g_1+2 f_1-2=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx%5E2&plus;y%5E2&plus;2%20g_1%20x&plus;2%20f_

If two circles \mathrm{x^2+y^2+2 g_1 x+2 f_1 y+2 g_1+2 f_1-2=0} and \mathrm{x^2+y^2+2 g_2 x+2 f_2 y+2 g_2+2 f_2-2=0} touch each other, then
 

Option: 1

\mathrm{g_1 f_2=g_2 f_1}

 


Option: 2

\mathrm{g_1^2 f_1=g_2 f_2^2}
 


Option: 3

\mathrm{\left(1-f_1\right)\left(1-g_2\right)=\left(1-g_1\right)\left(1-f_2\right)}
 


Option: 4

none of these


Answers (1)

best_answer

Point \mathrm{P(-1,-1)} satisfy both circles so they must touch at this point

\mathrm{p(-1,-1),C_1(-g_1,-f_1)andC_2(-g_2,-f_2)}are collinear points 

\mathrm{ Slope \: of\: \mathrm{P C_1= Slope \: of \: P C_2} }
\mathrm{ \Rightarrow \quad \frac{1-f_1}{1-g_1}=\frac{1-f_2}{1-g_2} \Rightarrow\left(1-f_1\right)\left(1-g_2\right)=\left(1-g_1\right)\left(1-f_2\right) }

Hence option 3 is correct

Posted by

manish painkra

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