Question

If the straight line y = mx + c touches the circle $x^2+y^2-4y=0,$ then the value of c will be

• A. $1+\sqrt{1+m^2}$
• B. $1-\sqrt{m^2+1}$
• C. $2(1+\sqrt{1+m^2})$
• D. $2+\sqrt{1+m^2}$

Answer 1

The correct option is C $2(1+\sqrt{1+m^2})$

Apply for tangency of line, centre being (0, 2) and radius = 2
$\left|\frac{-2+c}{\sqrt{1+m^2}}\right|=2\Rightarrow c^2-4c+4=4+4m^2$
$\Rightarrow c=\frac{4\pm \sqrt{16+16m^2}}{2}orc=2\pm 2\sqrt{1+m^2}$
Most correct answer is $c=2(1+\sqrt{1+m^2}).$