Question

The line $ (x-2)\cos \beta +(y-2)\sin \theta =1 $ touches a circle for all value of $ \theta , $ then the equation of circle is

• A. $ {{x}^{2}}+{{y}^{2}}-4x-4y+7=0 $
• B. $ {{x}^{2}}+{{y}^{2}}+4x+4y+7=0 $
• C. $ {{x}^{2}}+{{y}^{2}}-4x-4y-7=0 $
• D. None of the above

Answer 1

Answer: (A)

Give line is
$(x-2) \cos \theta+(y-2) \sin \theta=1=\cos ^{2} \theta+\sin ^{2} \theta$ On
comparing we get,
$x-2 \cos \theta \ldots$..(i)
$y-2=\sin \theta \ldots$...(ii)
and On squaring and then adding Eqs. (i) and (ii), we get
$(x-2)^{2}+(y-2)^{2}=\cos ^{2} \theta+\sin ^{2} \theta$
$\Rightarrow (x-2)^{2}+(y-2)^{2}=1 $
$\Rightarrow x^{2}+y^{2}-4 x-4 y+7=0$