Question

The equation of circle which touches $x \& y$ axis and whose perpendicular distance of centre of circle from $3 x+4 y+11=0$ is 5 is (Given that circle lies in lst quadrant)

• A. $x^2+y^2+4 x+4 y+4=0$
• B. $x^2+y^2-4 x-4 y+4=0$
• C. $x^2+y^2-4 x-4 y+8=0$
• D. $x^2+y^2-4 x-4 y-4=0$

Answer 1

Answer: (B)

Let centre be $(\alpha, \alpha) \& \alpha>0$
$\Rightarrow 5=\left|\frac{3 \alpha+4 \alpha+11}{5}\right| $
$\Rightarrow 7 \alpha+11=\pm 25$

$7 \alpha=14,7 \alpha=-36$
$\alpha=2, \alpha=\frac{-36}{7}$
so circle $(x-2)^2+(y-2)^2=4$