Circles in Polar Coordinates too

$Let\quad (R,\alpha )\quad be\quad the\quad coordinates\quad of\quad the\quad centre\quad of\quad the\quad circle\\and\quad let\quad a\quad be\quad its\quad radius,\\ then\quad its\quad equation\quad would\quad be\quad { r }^{ 2 }-2rRcos(\theta -\alpha )+({ R }^{ 2 }{ -\alpha }^{ 2 })=0\\ \\ \therefore \quad c=Rcos\alpha \quad and\quad a=Rsin\alpha .\\ \Rightarrow { \quad r }^{ 2 }-2r(ccos(\theta )+asin(\theta ))+({ c }^{ 2 })=0.\\ or\quad { \quad r }^{ 2 }-2rccos(\theta )+({ c }^{ 2 })=2arsin(\theta )....(i)\\ Since\quad it\quad passes\quad through\quad (r',\theta '),\quad we\quad have\\ { \quad r' }^{ 2 }-2r'ccos(\theta ')+({ c }^{ 2 })=2ar'sin(\theta ')....(ii)\\ (i)\div (ii)\quad \Rightarrow \quad \frac { { \quad r }^{ 2 }-2rccos(\theta )+({ c }^{ 2 }) }{ { \quad r' }^{ 2 }-2r'ccos(\theta ')+({ c }^{ 2 }) } =\frac { rsin(\theta ) }{ r'sin(\theta ') } \\ \Rightarrow \quad ({ r }^{ 2 }-2rccos(\theta )+({ c }^{ 2 }))=\frac { rsin(\theta ) }{ r'sin(\theta ') } ({ r' }^{ 2 }-2r'ccos(\theta ')+({ c }^{ 2 }))\\ \Rightarrow \quad \lambda \quad =\quad \frac { rsin(\theta ) }{ r'sin(\theta ') }$