Question

Prove that the equation $r^2\cos \theta -ar\cos 2\theta -2a2\cos \theta =0$ represents a straight line and a circle.

Answer 1

$r^2\cos \theta -ar\cos 2\theta -2a^2\cos \theta =0r^2\cos \theta -ar(2{\cos }^2\theta -1)-2a^2\cos \theta =0r^2\cos \theta -2ar{\cos }^2\theta +ar-2a^2\cos \theta =0r\cos \theta (r-2a\cos \theta )+a(r-2a\cos \theta )=0(r\cos \theta +a)(r-2a\cos \theta )=0\Rightarrow r\cos \theta +a=0,r-2a\cos \theta =0$

Converting $r\cos \theta +a=0$ to Cartesian coordinates

$\Rightarrow x+a=0$

which represents a straight line

Also $r-2a\cos \theta =0$

Multiplying by $r$ on both sides

$\Rightarrow r^2-2ar\cos \theta =0\Rightarrow {\left(\sqrt{x^2+y^2}\right)}^2-2ax=0\Rightarrow x^2+y^2-2ax=0$

which represents a circle .

Hence proved that the given equation represents a circle and a straight line