A) \[{{r}^{2}}={{c}^{2}}({{\lambda }^{2}}+{{\mu }^{2}})\]
B) \[{{c}^{2}}={{r}^{2}}({{\lambda }^{2}}+{{\mu }^{2}})\]
C) \[c=r({{\lambda }^{2}}+{{\mu }^{2}})\]
D) \[r=c({{\lambda }^{2}}+{{\mu }^{2}})\]
Correct Answer: B
Solution :
The equation of polar of the circle\[{{x}^{2}}+{{y}^{2}}+2\lambda x+2\mu y+c=0\]with respect to origin is \[x.0+y.0+\lambda (x+0)+\mu (y+0)+c=0\] \[\Rightarrow \] \[\lambda x+\mu y+c=0\] \[\Rightarrow \] \[y=-\frac{\lambda }{\mu }x-\frac{c}{\mu }\] If\[y=mx+c\]touches the circle\[{{x}^{2}}+{{y}^{2}}={{r}^{2}},\] then \[c=\pm \sqrt{1+{{m}^{2}}}\] Hence,\[y=-\frac{\lambda }{\mu }x-\frac{c}{\mu }\]will touch the circle \[{{x}^{2}}+{{y}^{2}}={{r}^{2}}\]if \[-\frac{c}{\mu }=\pm r\sqrt{1+{{\left( -\frac{\lambda }{\mu } \right)}^{2}}}\] \[\Rightarrow \] \[\frac{{{c}^{2}}}{{{\mu }^{2}}}={{r}^{2}}\left( \frac{{{\mu }^{2}}+{{\lambda }^{2}}}{{{\mu }^{2}}} \right)\] \[\Rightarrow \] \[{{c}^{2}}={{r}^{2}}({{\mu }^{2}}+{{\lambda }^{2}})\]
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