Question

The radius of the circle $2x^2+2y^2-4x\cos \theta +4y\sin \theta -1-4\cos \theta -\cos 2\theta =0$ is

• A. $1-\cos \theta$
• B. $1+\cos \theta$
• C. $1-\sin \theta$
• D. none of these

Answer 1

The correct option is B $1+\cos \theta$

$x^2+y^2-2x\cos \theta +2y\sin \theta -\frac{1}{2}[1+\cos 2\theta +4\cos \theta ]=0$

$x^2+y^2-2x\cos \theta +2y\sin \theta -\frac{1}{2}[2{\cos }^2\theta +4\cos \theta ]=0$

$x^2+y^2-2x\cos \theta +2y\sin \theta -{\cos }^2\theta -2\cos \theta =0$

$(x-\cos \theta {)}^2+(y+\sin \theta {)}^2-({\cos }^2\theta +{\sin }^2\theta )-{\cos }^2\theta -2\cos \theta =0$
Or
$(x-\cos \theta {)}^2+(y+\sin \theta {)}^2-1-{\cos }^2\theta -2\cos \theta =0$
Or
$(x-\cos \theta {)}^2+(y+\sin \theta {)}^2-(1+\cos \theta {)}^2=0$
Or
$(x-\cos \theta {)}^2+(y+\sin \theta {)}^2=(1+\cos \theta {)}^2$
Comparing with
$(x-\alpha {)}^2+(y-\beta {)}^2=r^2$ we get
$r=1+\cos \theta$.