Question

An equation of a circle touching the axes of coordinates and the line $x\cos \alpha +y\sin \alpha =2is{x}^2+y^2-2gx+2gy+g^2=0$ where $g=$

• A. $2{(\cos \alpha +\sin \alpha +1)}^{-1}$
• B. $2{(\cos \alpha -\sin \alpha +1)}^{-1}$
• C. $2{(\cos \alpha +\sin \alpha -1)}^{-1}$
• D. $-2{(\cos \alpha -\sin \alpha -1)}^{-1}$

Answer 1

The correct option is B $2{(\cos \alpha -\sin \alpha +1)}^{-1}$

The circle $x^2+y^2-2gx+2gy+g^2=0$ touches the co-ordinate axes.
Now, since it also touches the line $x\cos \alpha +y\sin \alpha =2$, the perpendicular distance from the centre to the line must be the radius.
Hence, $\frac{g\cos \alpha -g\sin \alpha -2}{\sqrt{{\cos }^2\alpha +{\sin }^2\alpha }}=\pm g$
$\Rightarrow g\cos \alpha -g\sin \alpha -2=\pm g$
$\Rightarrow g=\frac{2}{\cos \alpha -\sin \alpha +1}$ or $g=\frac{2}{\cos \alpha -\sin \alpha -1}$