Circles in Polar Coordinates

$\frac { 1 }{ r } =Acos\theta +Bsin\theta$ $\Rightarrow Arcos\theta +Brsin\theta =1\\ \Rightarrow Ax+By=1\\ as\quad radius\quad of\quad circle\quad is\quad a,\quad centre\quad lies\quad on\quad (a,0). \\Shifting\quad the\quad origin\quad to\quad the\quad centre\quad by\quad putting\quad x=X+a,\\ we\quad get\quad \\ y=-\frac { A }{ B } X+\frac { 1-Aa }{ B } \\ In\quad a\quad general\quad tangent\quad equation\quad y=mx+c \\ to\quad a\quad circle\quad x^{2}+y^{2} = {a}^{2} \\c=a\sqrt { 1+{ m }^{ 2 } } ,\\ \Rightarrow 1+{A}^2{ a }^{ 2 }-2Aa={ a }^{ 2 }{ B }^{ 2 }+{ a }^{ 2 }{ A }^{ 2 }\\ \therefore \quad { a }^{ 2 }{ B }^{ 2 }+2Aa=1.\\$