Question

Find the condition that straight line $\frac{k}{r}=A\cos \theta +B\sin \theta$ may touch the circle $r=2a\cos \theta$.

Answer 1

$x=r\cos \theta y=r\sin \theta r^2=x^2+y^2$

$k=A\left(r\cos \theta \right)+B\left(r\sin \theta \right)k=Ax+By$

$r=2a\cos \theta r^2=2a\left(r\cos \theta \right)x^2+y^2=2ax(x-a{)}^2+y^2=a^2$

Centre $(a,0)$ radius $=a$

Distance of centre from line $Ax+By-k=0$ must be equal to radius $a$

So that line $Ax+By-k$ would be a tangent to circle

$\left|\frac{Aa+B\left(0\right)-k}{\sqrt{A^2+B^2}}\right|=a|Aa-k|=a\sqrt{A^2+B^2}|Aa-k{|}^2=a^2(A^2+B^2)A^2{a}^2+k^2-2aAk=a^2{A}^2+B^2{a}^2{k}^2-2aAk-B^2{a}^2=0$