Question

The polar of the circle $ {{x}^{2}}+{{y}^{2}}+2\lambda x+2\mu y+c=0 $ with respect to origin will touch $ {{x}^{2}}+{{y}^{2}}={{r}^{2}}, $ if

• 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}}) $

Answer 1

Answer: (B)

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}}) $