Question

The line $(x-2)\cos \theta +(y-2)\sin \theta =1$ 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. $Noneoftheabove$

Answer 1

The correct option is A $x^2+y^2-4x-4y+7=0$

Given line is
$(x-2)cos\theta +(y-2)sin\theta =1$
$\Rightarrow (x-2)cos\theta +(y-2)sin\theta =co{s}^2\theta +si{n}^2\theta$
On compaining we get,
$(x-2)=cos\theta$ ..... $\left(i\right)$
$(y-2)=sin\theta$ ..... $\left(ii\right)$
On squaring and then adding Eqs. $\left(i\right)$ and $\left(ii\right),$ we get
$(x-2{)}^2+(y-2{)}^2=co{s}^2\theta +si{n}^2\theta$
$\Rightarrow (x-2{)}^2+(y-2{)}^2=1$
$\Rightarrow x^2+y^2-4x-4y+7=0$